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Applied Maths Chapter 18 Solutions
Circle And Parabola
EXERCISE- 18.1
Q.1 Find the equation of the circle whose :
(i) centre is (0, 2) and radius 2
(ii) centre is (-3, 2) and radius 4
(iii) The centre is (1/2, 1/4) and radius 1/12
(iv) centre is (-a,-b) and radius √a^2 – b^2.
Ans.
(i) centre is (0, 2) and radius 2
(x – h)^2 + (y – k)^2 = r^2 [Central form]
(x – 0)^2 + (y – 2)^2 = (2)^2
x^2 + y^2 – 4y + 4 = 4
x^2 + y^2 – 4y = 0
(ii) centre is (-3, 2) and radius 4
(x – h)^2 + (y – k)^2 = r^2 [Central form]
(x + 3)^2 + (y – 2)^2 = (4)^2
x^2 + 9 + 6x + y^2 + 4 – 4y = 16
x^2 + y^2 + 6x – 4y +13 – 16 = 0
x^2 + y^2 + 6x – 4y -3 = 0
(iii) The centre is (1/2, 1/4) and radius 1/12
(x – h)^2 + (y – k)^2 = r^2 [Central form]
(x – 1/2)^2 + (y – 1/4)^2 = (1/12)^2
x^2 + 1/4 – x + y^2 + 1/16 – y/2 = 1/144
x^2 + y^2 – x – y/2 +1/4 + 1/16 – 1/144 = 0
(144x^2 + 144y^2 – 144x – 72y + 36 + 9 – 1)/144 = 0
144x^2 + 144y^2 – 144x – 72y + 44 = 44 0
4 (36x^2 + 36y^2 – 36x – 18y + 11) = 0
Therefore, 36x^2 + 36y^2 – 36x – 18y + 11 = 0
(iv) centre is (-a,-b) and radius √a^2 – b^2.
(x – h)^2 + (y – k)^2 = r^2 [Central form]
(x – a)^2 + (y – b)^2 = (√a^2 – b^2)^2
x^2 + a^2 + 2ax + y^2 + b^2 + 2by = 0
x^2 + y^2 + 2ax + 2by + 2b^2 = 0
Q.2 Determine the equation of a circle whose centre is (8, -6) and which passes through the point (5,-2).
Ans. (x – h)^2 + (y – k)^2 = r^2…………(i)
(5 – 8)^2 + (-2 + 6)^2 = r^2
(-3)^2 + (4)^2 = r^2
9 + 16 = r^2
r^2 = 25
r = 5
from eq (i)
(x – h)^2 + (y – k)^2 = r^2
(x – 8)^2 + (y + 6)^2 = (5)^2
x^2 + 64 – 16x + y^2 + 36 + 12y = 25
x^2 + y^2 – 16x + 12y + 75 = 0
Q.3 Find the equation of the circle whose centre lies on the negative direction of the y-axis at a distance of 3 units from the origin and whose radius is 4 units.
Ans. Point (0, 3), r = 4
(x – h)^2 + (Y – k)^2 = r^2
(x – 0)^2 + (y + 3)^2 = (4)^2
x^2 + y^2 + 6y + 9 = 16
x^2 + y^2 + 6y – 7 = 0
Q.4 Find the equation of the circle which has the points (-2, 3) and (0, -1) as opposite ends of a diameters.
Ans. (x – x1) (x – x2) + (y – y1) (y -y2) = 0 [Diameter form]
(x + 2) (x -0) + (y – 3) (y + 1) = 0
x^2 + 2x + y^2 + y – 3y – 3= 0
x^2 + y^2 + 2x – 2y – 3 = 0
Q.5 (i) Does the point (-2.5, 3.5) lie inside, outside or on the circle x^2 + y^2 = 25?
(ii) Does the (-3, 7) lie inside, outside or on the circle x^2 + y^2 = 49?
Ans.
(i) Does the point (-2.5, 3.5) lie inside, outside or on the circle x^2 + y^2 = 25?
x^2 + y^2 = 25
x^2 + y^2 = (5)^2
Therefore, r = 5
center (0, 0)
Using Distance Formula,
OP = √(2.5 – 0)^2 + (3.5 – 0)^2
= √6.25 + 12.25
= √18.25
= 4.2 < 5
Hence, P lies inside the circle
(ii) Does the (-3, 7) lie inside, outside or on the circle x^2 + y^2 = 49?
x^2 + y^2 = 49
x^2 + y^2 = (7)^2
Therefore, r = 7
center (0, 0)
Using Distance Formula,
OP = √(-3 – 0)^2 + (7 – 0)^2
= √9 + 49
= √58
= 7.6 > 7
Hence, P lies outside the circle
Q.6 Find the centre and the radius of the circle:
(i) (x + 5)^2 + (y – 3)^2 = 36
(ii) x^2 + y^2 – 4x – 8y – 45 = 0
(iii) 2x^2 + 2y^2 – 3x + 5y – 7 = 0
(iv) x^2 + y^2 – ax – by= 0.
Ans.
(i) (x + 5)^2 + (y – 3)^2 = 36
(x – h)^2 + (y – k)^2 = r^2 [Central form]
(x – (5))^2 + (y – 3)^2 = (6)^2
h = -5, k = 3, r = 6
Centre (-5, 3), r = 6
(ii) x^2 + y^2 – 4x – 8y – 45 = 0
x^2 + y^2 + 2gx + c = 0 [General form]
2g = -4 , 2f = -8 , c = -45
g = -2 , f = -4
Centre (-g, -f) = (2, 4)
r = √g^2 + f^2 – c
= √4 + 16 – (-45)
= √65
(iii) 2x^2 + 2y^2 – 3x + 5y – 7 = 0
x^2 + y^2 + 2gx + c = 0 [General form]
1/2 (2x^2 + 2y^2 – 3x + 5y -7) = 0
x^2 + y^2 – 3/2x + 5/2y – 7/2 = 0
2g = -3/2 , 2f = 5/2 , c = -7/2
g = -3/4 , f = 5/4
Centre (-g, -f) = (3/4, -5/4)
r = √g^2 + f^2 – c
= √9/16 + 25/16 – (-7/2)
= √(9 + 25 + 56)/16
= √93/16
= 3/4√10
7. Which of the following equations represents a circle? If so, determine its centre and radius:
(i) x^2 + y^2 + x – y = 0
(ii) x^2 + y^2 – 3x +3y + 10 = 0
(iii) 2x^2 + 2y^2 = 5x + 7y + 3
(iv) x^2 + y^2 + 2x + 10y + 26 = 0.
Ans. (i) x^2 + y^2 + x – y = 0
2g = 1, 2f = -1, c = 0
g = 1/2, f = -1/2
g^2 + f^2 – c > 0
(1/2)^2 + (-1/2)^2 – 0 > 0
1/4 + 1/4 > 0
1/2 > 0
Therefore, it is equation of circle.
Centre (-1/2, 1/2), r = √g^2+f^2-c
= √1/2 = 1/√2
(ii) x^2 + y^2 – 3x +3y + 10 = 0
2g = -3, 2f = 3, c = 10
g = -2/3, f = 2/3, c = 10
g^2 + f^2 – c > 0
(-2/3)^2 + (2/3)^2 – 10 > 0
9/4 + 9/4 -10 > 0
(9 + 9 – 40)/4 >0
-22/4 < 0
Therefore, it is not a circle.
(iii) 2x^2 + 2y^2 = 5x + 7y + 3
x^2 + y^2 – 5/2x – 7/2y – 3/2 = 0
2g = -5/2, 2f = -7/2, c = -3/2
g = -5/4, f = -7/4, c = -3/2
g^2 + f^2 – c >0
(-5/4)^2 + (-7/4)^2 – 3/2 > 0
25/16 + 49/16 + 3/2 > 0
(25 + 49 + 24)/16 > 0
98/16 > 0
Therefore, it is equation of circle.
Centre (5/4, 7/4), r = √g^2+f^2-c
= √98/16 = 7/2√2
Q.8 Find the value of p so that x^2 + y^2+ 8x + 10y + p = 0 is the equation of a circle of radius 7 units.
Ans. x^2 + y^2+ 8x + 10y + p = 0
2g = 8, 2f = 10
g = 4, f = 5
Using, √g^2+f^2-c = r
√4^2+5^2-p = 7
16 + 25 – p = 7
41 – 49 = p
p = -8
Q.9 (i) Find the equation of a circle concentric with the circle x^2 + y^2 – 8x + 2y + 3 = 0 and of radius 3 units.
(ii) Find the equation of the circle which passes through the point (1,-2) and is concentric with the circle x^2 + y^2 – 4x + 5y – 7 = 0.
Ans. (i) x^2 + y^2 – 8x + 2y + 3 = 0
2g = -8, 2f = 2
g = -4, f = 1
Centre (4, -1)
Equation of circle, (x – h)^2 + (y – k)^2 = r^2
(x – 4)^2 + (y + 1)^2 = 3^2
x^2 – 8x + 16 + y^2 + 2y + 1 = 9
x^2 + y^2 – 8x + 2y + 8 = 0
(ii) x^2 + y^2 – 4x + 5y – 7 = 0
2g = 4, 2f = 5
g = -2, f = 5/2
Centre (2, -5/2)
Using, (x – h)^2 + (y – k)^2 = r^2
(1 – 2)^2 + (-2 + 5/2)^2 = r^2
1 + 1/4 = r^2
r^2 = 5/4
Equation of circle, (x – h)^2 + (y – k)^2 = r^2
(x – 2)^2 + (y + 5/2)^2 = 5/4
x^2 – 4x + 4 + y^2 + 5y + 25/4 – 5/4 = 0
x^2 + y^2 – 4x + 5y + 9 = 0
Q.10 Find the equation of the circle which passes through the point (2, 4) and centre at the intersection of the lines x – y = 4 and 2x + 3y + 7 = 0.
Ans. The center of the circle is at the intersection of the two given lines:
- x − y = 4 ………….(i)
- 2x + 3y + 7 = 0…..(ii)
By using elimination method,
( x – y = 4 ) x 2
2x – 2y = 8
2x + 3y = -7
(-) (-) (+)
-5y = 15
y = -3
From eq. (i) x – y = 4
x – (-3) = 4
x = 4 – 3
x = 1
Point (1, -3)
Therefore, centre of circle (1, -3)
The general equation of a circle with center (h ,k) and radius r is :
(x − h)^2 + (y − k)^2 = r^2
(2 – 1)^2 + (4 + 3)^2 = r^2
1 + 49 = r^2
r^2 = 50
Equation of circle, (x + 1)^2 + (y – 3)^2 = 50
x^2 – 2x + 1 + y^2 + 6y + 9 – 50 = 0
x^2 + y^2 – 2x + 6y – 40 = 0
Q.11 If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 sq. units, then obtain the equation of the circle.
Ans. The center of the circle is at the intersection of the two lines, which are given as diameters of the circle.
- 2x − 3y = 5
- 3x − 4y = 7
Multiply the first equation by 3 and the second by 2 to eliminate x :
6x − 9y = 15………(Equation 1)
6x − 8y = 14………(Equation 2)
By using elimination method :
6x – 9y = 15
6x – 8y = 14
(-) (+) (-)
-y = 1
y = 1
Substitute y = −1 back into the first original equation to find x :
2x − 3(−1) = 5
2x = 2 ⇒ x = 1
So, the center of the circle is at (1,−1).
Area = πr2
154 = 22/7 x r^2
(154 x 7)/22 = r^2
r = 7
The general equation of a circle with center (h, k) and radius r is :
(x−h)^2 + (y−k)^2 = r^2
Substituting h = 1, k = −1, and r = 7 :
(x – 1)^2 + (y + 1)^2 = 7^2
x^2 – 2x + 1 + y^2 + 2y + 1 = 49
x^2 + y^2 – 2x + 2y – 47 = 0
Q.12 (i) Find the equation of the circle of radius 5 units whose centre lies on the y-axis and passes through (3,2).
(ii) Find the equations of circles which pass through two points on the x-axis at distances of 4 units from the origin and whose radius is 5 units.
Ans. (i) To find the equation of a circle with a radius of 5 units, whose center lies on the y-axis and passes through the point (3,2),
Centre lies on y-axis
Therefore, Centre (0, k)
Using, (x – h)^2 + (y – k)^2 = r^2
(3 – 0)^2 + (y – k)^2 = 5^2
9 + 4 – 4k + k^2 = 25
k^2 – 4k – 12 = 0
(k – 6) (k + 2) = 0
k = 6, k = -2
If k = 6,
(x – 0)^2 + (y – 6)^2 = 5^2
x^2 + y^2 – 12y + 36 = 25
x^2 + y^2 – 12y + 11 = 0
If k = -2
(x – 0)^2 + (y + 2)^2 = 5^2
x^2 + y^2 + 4y + 4 = 25
x^2 + y^2 + 4y – 21 = 0
(ii) To find the equations of circles that pass through two points on the x-axis at distances of 4 units from the origin and have a radius of 5 units,
AP^2 = OP^2 + OA^2
5^2 = OP^2 + 4^2
25 – 16 = OP^2
OP^2 = 9
OP = 3
Center of circle (0, 3)
(x – 0)^2 + (y – 3)^2 = 25
x^2 + y^2 – 6y + 9 – 25 = 0
x^2 + y^2 – 6y – 16 = 0
If, Centre (0, -3)
Then eq, x^2 + y^2 + 6y – 16 = 0
Q.13 (i) Find the equation of the circle which has A(1, 3) and B (4, 5) as opposite ends of a diameter. Find also the equation of the perpendicular diameter.
(ii) Find the centre, radius and the equation of the circle drawn on the line segment joining the points (-1, 2) and (3,-4) as diameter.
Ans. (i) To find the equation of the circle with points A(1,3) and B(4,5) as opposite ends of a diameter,
Equation of circle,
(x – 1)(x – 4) + (y – 3)(y – 5) = 0
x^2 – 4x + x + 4 + y^2 – 5y – 3y + 15 = 0
x^2 + y^2 – 5x – 8y + 19 = 0
The perpendicular diameter will be the line passing through the center of the circle and perpendicular to the line AB.
The slope of AB is given by:
Slope of AB = (y2−y1)/(x2 − x1)
= (5 -3)/(4 – 1) = 2/3
The slope of the line perpendicular to AB will be the negative reciprocal of 2/3, which is -3/2.
Therefore, m = -3/2
The center of the circle is the midpoint of the line segment AB. The formula for the midpoint (x, y) between two points (x1, y1) and (x2, y2) is:
[(x1 + x2)/2,(y1 + y2)/2]
Applying this formula to points A(1, 3) and B(4, 5):
Centre [(1 + 4)/2, (3 + 5)/2] = (5/2, 4)
Therefore, CD passes through (5/2, 4)
Equation of CD : (y – y1) = m (x – x1)
(y – 4) = -3/2 (x – 5/2)
2y – 8 = -3x + 15/2
4y – 16 = -6x + 15
6x + 4y – 31 = 0
Q.14 Find the equation of the circle drawn on a diagonal of the rectangle as its diameter whose sides are the lines given by: x = 4, x = -2, y = 5 and y = -2.
Ans. x = 4, x = -2, y = 5 and y = -2
The vertices of the rectangle are:
- (4,5)
- (4,−2)
- (−2,5)
- (−2,−2)
The diagonals of the rectangle are:
- One diagonal connects (4,5) and (−2,−2)
- The other diagonal connects (4,−2) and (−2,5)
Let’s consider the diagonal connecting (4,5) and (−2,−2) as the diameter of the circle.
Equation of circle : (x – x1) (x – x2) + (y – y1) (y – y2) = 0
(x – 4) (x + 2) + (y – 5) (y + 2) = 0
x^2 + 2x – 4x – 8 + y^2 + 2y – 5y – 10 = 0
x^2 + y^2 – 2x – 3y – 18 = 0
Q.15 Find the equation of the circle drawn on the intercept made by the line 3x + 4y = 12 between the coordinate axes as diameter.
Ans. The intercepts can be found by setting y = 0 to find the x-intercept and setting x = 0 to find the y-intercept.
- x-intercept: Set y = 0 in the equation of the line: 3x + 4(0) = 12 ⟹ 3x = 12 ⟹ x = 4
- y-intercept: Set x = 0 in the equation of the line: 3(0) + 4y = 12 ⟹ 4y = 12 ⟹ y = 3
The diameter of the circle is the line segment joining the points (4,0) and (0,3)
Equation of circle : (x – x1) (x – x2) + (y – y1) (y – y2) = 0
(x – 4) (x – 0) + (y – 0) (y – 3) = 0
x^2 – 4x + y^2 – 3y = 0
x^2 + y^2 – 4x – 3y = 0
Q.16 Find the equation of the circle circumscribing a square ABCD with sides AB and AD as coordinate axes.
Ans. To find the equation of the circle circumscribing a square ABCD with sides AB and AD as coordinate axes,
Given that sides AB and AD of the square are along the coordinate axes, we can place the square in the coordinate plane as follows:
- Point A is at the origin (0,0).
- Point B lies on the positive x-axis, so let’s denote B as (a,0).
- Point D lies on the positive y-axis, so let’s denote D as (0,a).
- Point C would then be diagonally opposite to A and hence would be at (a, a).
Equation of circle, (x – a) (x – 0) + (y – 0) (y – a) = 0
x^2 – ax + y^2 – ay = 0
x^2 + y^2 – ax – ay = 0
x^2 + y^2 – a(x + y) = 0
Q.17 The radius of the circle x^2 + y^2 – 2x + 3y + k = 0 is 22. Find the value of k. Find also the equation of the diameter of the circle which passes through the point (5, 2).
Ans. To find the value of k and the equation of the diameter of the circle that passes through the point (5, 2), let’s first rewrite the given equation of the circle in standard form.
x^2 + y^2 – 2x + 3y + k = 0
2g = -2, 2f = 3, c = k, r = 2 1/2 = 5/2
g = -1, f = 3/2
r = √g^2+f^2-c
5/2 = √1+9/4-k
25/4 = 13/4 – k
k = 13/4 – 25/4
k = -12/4
k = -3
Equation of diameter :
(y – y1) = (y2 – y1)/(x2 – x1) (x – x1)
(y + 3/2) = [(5/2)+(3/2)]/(5-1) (x – 1)
(2y + 3)/2 = 4/4 (x – 1)
2y + 3 = 2x – 2
2x – 2y – 5 = 0
Q.18 Find the equation of the circle the endpoints of whose one diameter are the centre of the circles x^2 + y^2 + 6x – 14y + 5 = 0 and x^2 + y^2 – 4x + 10y + 7 = 0.
Ans. To find the equation of the circle whose endpoints of one diameter are the centers of the given circles,
x^2 + y^2 + 6x – 14y + 5 = 0
2g = 6, 2f = -14
g = 3, f = -7
Centre (-g, -f) = (-3, 7)
x^2 + y^2 – 4x + 10y + 7 = 0
2g = -4, 2f = 10
g = -2, f = 5
Equation of circle,
(x + 3)(x – 2) + (y – 7)(y + 5) = 0
x^2 – 2x + 3x – 6 + y^2 + 5y – 7y – 35 = 0
x^2 + y^2 + x – 2y – 41 = 0
Q.19 One end of a diameter of the circle x^2 + y^2 – 3x + 5y – 4 = 0 is (2, 1), find the coordinates of the other end.
Ans. To find the coordinates of the other end of the diameter of the circle, we first need to determine the center of the circle. The general equation of a circle is given by:
x^2 + y^2 + 2gx + 2fy + c = 0
2g = -3, 2f = 5
g = -3/2, f = 5/2
Centre (-g, -f)
(32, -5/2)
If (x1, y1) = (2, 1) is one end of the diameter, and (x2, y2) is the other end, the midpoint of the diameter is the center of the circle. Using the midpoint formula, we have:
x = (x1 + x2)/2, y = (y1 + y2)/2
3/2 = (2 + x2)/2, -5/2 = (1 + y2)/2
x2 = 1 , y2 = -6
Therefore, coordinate of B (1, -6)
Q.20 Find the equation of the circle concentric with the circle x^2 + y^2 – 4x – 6y – 9 = 0 and passing through the point (-4,-5).
Ans. To find the equation of a circle that is concentric with another circle and passes through a specific point, we need to determine two things: the center of the concentric circle and its radius.
2g = -4, 2f = -6
g = -2, f = -3
Centre (2, 3)
Equation of a circle = (x -h)^2 + (y – k)^2 = r^2
(-4 – 2)^ + (-5 – 3)^2 = r^2
36 + 4 = r^2
r^2 = 100
r = 10
So, (x – 2)^2 + (y – 3)^3 = 100
x^2 – 4x + 4 + y^2- 6y + 9 – 100 = 0
x^2 + y^2 – 4x – 6y – 87 = 0
Q.21 Find the equation of the circle which passes through the centre of the circle x^2 + y^2 + 8x + 10y + 7 = 0 and is concentric with the circle 2x^2 + 2y^2 – 8x – 12y – 9 = 0.
Ans.
Q.22 Find the equation of the circle which is concentric with the circle x^2 + y^2 – 4x + 6y – 3 =0 and of double its
(i) circumference
(ii) area.
Ans. x^2 + y^2 – 4x + 6y – 3 = 0
2g = -4, 2g = 6, c = -3
g = -2, f = 3, c = -3
r = √g^2 + f^2 – c
= √(-2)^2 + (3)^2 + 3
= √4 + 9 + 3
= √16
= 4
(i) Circumference = 2π x 4 = 8π
Area = πr^2
= π(4)^2 = 16π
If circumference = 16π
2π R = 16π
R = 8
Equation of circle
(x – h)^2 + (y – k)^2 = r^2
(x – 2)^2 + (y – 3)^2 = 8^2
x^2 – 4x + 4 + y^2 + 6y + 9 = 64
x^2 + y^2 – 4x + 6y – 51 = 0
(ii) If Area = 16 x 2π
= 32π
πR^2 = 32π
R^2 = 32
Equation of circle
(x – h)^2 + (y – k)^2 = r^2
(x – 2)^2 + (y – 3)^2 = 32
x^2 – 4x + 4 + y^2 + 6y + 9 – 32 = 0
x^2 + y^2 – 4x + 6y – 19 = 0
Q.23 Prove that the radii of the circles x^2 + y^2 = 1, x^2 + y^2 – 2x – 6y – 6 = 0 and x^2 + y^2 – 4x – 12y – 9 = 0 are in A.P.
Ans. x^2 + y^2 = (1)^2
Here, radius (r1) = 1
x^2 + y^2 – 2x – 6y – 6 = 0
2g = -2, 2g = -3, c = -6
g = -1, f = -6, c = -6
r2 = √g^2 + f^2 – c
= √(-1)^2 + (-3)^2 + 6
= √1 + 9 + 6
= √16
= 4
x^2 + y^2 – 4x – 12y – 9 = 0
2g = -4, 2g = -12, c = -9
g = -2, g = – 6, c = -9
r3 = √g^2 + f^2 – c
= √(-2)^2 + (-6)^2 + 9
= √4 + 36 + 9
= √49
= 7
Because common difference is the same
Therefore, it is an AP.
Q.24 (i) Find the equation of the circle passing through the points (4, 1), (6, 5) and whose centre is on the line 4x + y = 16.
(ii) Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the line y -4x + 3 = 0.
Ans. (i) Equation of a circle
(x -h)^2 + (y – k)^2 = r^2
as it passes through (4, 1)
(4 – h)^2 + (1 – k)^2 = r^2…….(i)
The second point passes through (6, 5)
(6 – h)^2 + (5 – k)^2 = r^2…….(ii)
The Centre of the circle lies on 4h + k = 16……(iii)
from eq. (i) & (ii)
(4 – h)^2 + (1 – k)^2 = (6 – h)^2 + (5 – k)^2
16 + h^2 – 8h + 1 + k^2 – 2k = 36 + h^2 – 12h + 25 + k^2 -10k
4h + 8k = 44……..(iv)
from eq. (iii) & (iv) [using elimination method]
4h + k = 16
4h + 8k = 16
(-) (-) (-)
-7k = -28
k = 4
By putting the value of k in eq.(i)
4h + k = 16
4h + 4 = 16
h = (16 – 4)/4
= 12/4 = 3
Q.25 Find the equation of the circle passing through the three points
(i) (1, 2), (3,-4) and (5, -6)
(ii) (20, 3), (19, 8) and (2,-9).
Also, find its centre and radius.
Ans. To find the equation of a circle passing through three given points, we can use the equation of a circle: (x – h)^2 + (y – k)^2 = r^2
Taking points (1, 2) = (1 – h)^2 + (2 – k)^2 = r^2…….(i)
Taking points (3, -4) = (3 – h)^2 + (-4 – k)^2 = r^2…(ii)
Taking points (5, -6) = (5 – h)^2 + (-6 – k)^2 = r^2….(iii)
From eq. (i) & (ii)
(1 – h)^2 + (2 – k)^2 = (3 – h)^2 + (-4 – k)^2
1 – 2h + h^2 + 4 – 4k + k^2 = 9 – 6h + h^2 + 16 + 8k + k^2
4h – 12k = 20
h – 3k = 5…….(iv)
From eq. (ii) & (iii)
(3 – h)^2 + (-4 – k)^2 = (5 – h)^2 + (-6 – k)^2
9 – 6h + h^2 + 16 + 8k + k^2 = 25 – 10h + h^2 + 36 + 12h + k^2
4h – 4k = 36
h – k = 9……….(v)
From (iv) & (v)
h – 3k = 5
h – k = 9
(-) (+) (-)
-2k = -4
k = 2
So, h = 11
From eq. (i) (1 – h)^2 + (2 – k)^2 = r^2
(1 – 11)^2 + (2 – 2)^2 = r^2
r^2 = 100
Therefore, eq. of circle
(x – 11)^2 + (y – 2)^2 = 100
x^2 – 22x + 121 + y^2 – 4y + 4 – 100 = 0
x^2 + y^2 – 22x – 4y + 25 = 0
Hence, Centre (11, 2), r = 10
Q.26 Find the equation of the circle which is circumscribed about the triangle whose vertices are (-2,3), (5, 2) and (6, -1).
Ans.
Q.27 Show that the points (7, 1), (-2, 4), (5, 5) and (6, 4) are concyclic. Also find the equation, centre and radius of the circle on which they lie.
Ans. To show that the points (7, 1), (−2, 4), (5, 5) and (6, 4) are concyclic, we need to demonstrate that they all lie on a single circle. This can be done by finding the equation of the circle passing through any three of these points and then verifying that the fourth point satisfies this equation.
equation of a circle: (x – h)^2 + (y – k)^2 = r^2
Taking points (7, 1) = (7 – h)^2 + (1 – k)^2 = r^2…….(i)
Taking points (-2, 4) = (-2 – h)^2 + (4 – k)^2 = r^2…(ii)
Taking points (5, 5) = (5 – h)^2 + (5 – k)^2 = r^2….(iii)
From eq. (i) & (ii)
(7 – h)^2 + (1 – k)^2 = (-2 – h)^2 + (4 – k)^2
49 – 14h + h^2 + 1 – 2k + k^2 = 4 – 4h + h^2 + 16 – 8k + k^2
18h – 6k = -30
3h – k = 5…….(iv)
From eq. (ii) & (iii)
(-2 – h)^2 + (4 – k)^2 = (5 – h)^2 + (5 – k)^2
4 + 4h + h^2 + 16 – 8k + k^2 = 25 – 10h + h^2 + 25 – 10h + k^2
14h + 2k = 30
7h + k = 15……….(v)
From (iv) & (v)
3h – k = 5
7h + k = 15
10h = 20
h = 2
So, k = 1
From eq. (i) (7 – h)^2 + (1 – k)^2 = r^2
(7 – 2)^2 + (1 – 1)^2 = r^2
r^2 = 25
r = 5, Centre (2, 1)
Now, eq. of circle
(x – 2)^2 + (y – 1)^2 = 25
x^2 – 4x + 4 + y^2 – 2y + 1 – 25 = 0
x^2 + y^2 – 4x – 2y – 20 = 0
FAQ’s related to Applied Maths Chapter 18 on Circle And Parabola:
Q.1 What is the equation of a circle and how is it derived?
Ans. The equation of a circle is (𝑥−ℎ)^2 + (𝑦−𝑘)^2= r^2, where (ℎ, 𝑘)(h, k) is the center of the circle and 𝑟 is the radius. This equation is derived from the distance formula between two points.
Q.2 What are the standard forms of the equation of a parabola?
Ans. The standard forms of the equation of a parabola are:
- Vertical Axis: 𝑦 = 𝑎𝑥^2 + 𝑏𝑥 + c
- Horizontal Axis: 𝑥 = 𝑎𝑦^2 + 𝑏𝑦 + 𝑐
These are few Frequently Asked Questions relating to Applied Maths Chapter 18
In Applied Maths chapter 18, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 18 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 18, we delve deep into advanced mathematical concepts that are crucial for understanding.
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