11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.1 Questions 15
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Question (1)

Find the equation of the circle with centre (0, 2) and radius 2

Solution

The equation of a circle with centre (h, k) and radius r is given as (x . h)2 + (y . k)2 = r2 It is given that centre (h, k) = (0, 2) and radius (r) = 2. Therefore, the equation of the circle is (x – 0)2 + (y – 2)2 = 22 x2 + y2 + 4 – 4 y = 4 x2+ y2 – 4y = 0

Question (2)

Find the equation of the circle with centre (–2, 3) and radius 4

Solution

The equation of a circle with centre (h, k) and radius r is given as (x. h)2 + (y . k)2 = r2 It is given that centre (h, k) = (–2, 3) and radius (r) = 4. Therefore, the equation of the circle is (x + 2)2 + (y – 3)2= (4)2 x2 + 4x + 4 + y2 – 6y + 9 = 16 x2 + y2 + 4x – 6y – 3 = 0

Question (3)

Find the equation of the circle with centre $\left( {\frac{1}{2},\frac{1}{4}} \right)$ and radius $\frac{1}{{12}}$

Solution

The equation of a circle with centre (h, k) and radius r is given as (x. h)2 + (y – k)2 = r2
It is given that centre $\left( {h,k} \right) = \left( {\frac{1}{2},\frac{1}{4}} \right)$ and radius $\left( r \right) = \frac{1}{{12}}$
Therefore, the equation of the circle is ${\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{4}} \right)^2} = {\left( {\frac{1}{{12}}} \right)^2}$
${x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} = \frac{1}{{144}}$
${x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} - \frac{1}{{144}} = 0$
144x2 - 144x +36+144y2 - 72y +9-1 = 0
144x2 - 144x +144y2 - 72y + 44 = 0
36x2 - 36x +36y2 - 18y + 11 = 0
36x2 +36y2 - 36x - 18y + 11 = 0

Question (4)

Find the equation of the circle with centre (1, 1) and radius √2

Solution

The equation of a circle with centre (h, k) and radius r is given as (x.h)2 + (y . k)2 = r2
It is given that centre (h, k) = (1, 1) and radius (r) = √2.
Therefore, the equation of the circle is
(x-1)2 + (y-1)2 = (√2)2
x2 - 2x + 1 + y2 - 2y +1 = 2
x2 + y2 - 2y - 2x = 0

Question (5)

Find the equation of the circle with centre (–a, –b) and radius $\sqrt {{a^2} - {b^2}}$

Solution

The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (–a, –b) and radius $\left( r \right) = \sqrt {{a^2} - {b^2}}$
Therefore, the equation of the circle is
${\left( {x + a} \right)^2} + {\left( {y + b} \right)^2} = {\left( {\sqrt {{a^2} - {b^2}} } \right)^2}$
x2 +2ax +a2 + y2 + 2by + b2 = a2 - b2
x2 + y2 + 2ax +2by +2b2 = 0

Question (6)

Find the centre and radius of the circle (x + 5)2 + (y – 3)2 = 36

Solution

The equation of the given circle is (x + 5)2 + (y – 3)2 = 36.
(x + 5)2 + (y – 3)2 = 36
⇒ {x – (–5)}2 + (y – 3)2 = 62, which is of the form (x – h)2 + (y – k)2 = r2, where h = –5, k = 3, and r = 6.
Thus, the centre of the given circle is (–5, 3), while its radius is 6.

Question (7)

Find the centre and radius of the circle x2 + y2 – 4x – 8y – 45 = 0

Solution

Given equation for circle x2 + y2 – 4x – 8y – 45 = 0
Comparing it to standard form x2 + y2 + 2gx + 2fy + c = 0 we get
g = -2, f = -4 and c = 45
centre = (-g, -f) = (2, 4)
radius $\left( r \right) = \sqrt {{g^2} + {f^2} - c}$
$\left( r \right) = \sqrt {4 + 16 + 45} = \sqrt {65}$

Question (8)

Find the centre and radius of the circle x2 + y2 – 8x + 10y – 12 = 0

Solution

Given equation for circle x2 + y2 – 8x + 10y – 12 = 0
Comparing it to standard form x2 + y2 + 2gx + 2fy + c = 0 we get
g = -4, f = 5 and c = -12
centre C(-g, -f) = (4, -5)
radius $\left( r \right) = \sqrt {{g^2} + {f^2} - c}$
$\left( r \right) = \sqrt {16 + 25 + 12} = \sqrt {53}$

Question (9)

Find the centre and radius of the circle 2x2 + 2y2 – x = 0

Solution

Given equation for circle is 2x2 + 2y2 – x = 0
or ${x^2} + {y^2} - \frac{1}{2}x = 0$
Comparing it to standard form x2 + y2 + 2gx + 2fy + c = 0 we get
centre C(-g, -f) = $\left( {\frac{1}{4},0} \right)$
radius $\left( r \right) = \sqrt {{g^2} + {f^2} - c}$
$\left( r \right) = \sqrt {\frac{1}{{16}} + 0 + 0} = \frac{1}{4}$

Question (10)

Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Solution

circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0
∴ centre of Circle, c = (-g, -f)
c∈ line 4x+y=16
∴ -4g -f - 16 = 0 ---(1)
(4,1) belongs to circle
∴ 42 + (1)2 + 2g(4) + 2f(1) + c =
16+ 1 + 8g + 2f + c = 0
8g + 2f + c +17 = 0 ---(2)
(6, 5) ∈ circle
∴ 36 +25 +2g(6) +2 f(5) + c = 0
12g + 10f + c + 61 = 0
---(3)
(2) - (3)
⇒ -4g - 8f - 44 = 0 ---(4)
(1)-(4)
⇒ 7f + 28 = 0
f = -4
Substitute f = -4 in (1) we get
-4g + 4 -16 = 0
-4g = 12
g = -3
Substitute f = -4 and g = -3 in (2)
8(-3) + 2(-4) + c + 17 = 0
-24-8+c 17 = 0
c = 15
∴ equation of circle will be
x2 + y2 - 6x - 8y +15 = 0

Question (11)

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

Solution

circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.
Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0
centre of circle c = (-g, -f)
c ∈ x - 3y-11 = 0
∴ -g+3f - 11 = 0 ---(1)
(2, 3) ∈ circle
22 + 32 +2g(2) +2f(3) + c = 0
4g + 6f +c +13 = 0 ---(2)
(-1, 1) ∈ circle
(-1)2 + (1)2 + 2g(-1) +2f(1) + c = 0
-2g +2f + c + 2 = 0 ---(3)
(2) - (3)
6g + 4f + 11 = 0 ----(3A)
6(1) + (3A)
-6g +18f -66 + 6g +4f + 11 = 0
22f = 55
$f = \frac{5}{2}$
Replace f in (1)
$- g + \frac{{15}}{2} - 11 = 0$
$g = - \frac{7}{2}$
Replace $g = - \frac{7}{2} \;and \;f = \frac{5}{2}$ in (1)
$4\left( { - \frac{7}{2}} \right) + 6\left( {\frac{5}{2}} \right) + c + 13 = 0$
c - 14 +15 + 13 = 0
c = -14
∴ Equation of circle will be
${x^2} + {y^2} + 2\left( {\frac{{ - 7}}{2}} \right)x + 2\left( {\frac{5}{2}} \right)y - 14 = 0$
x2 + y2 -7x + 5y - 14 = 0

Question (12)

Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Solution

r=5, c ∈ x-axis, passing through (2, 3) Since centre ∈ x-axis
Let p(2,3) ∈ circle
∴ cp = r
∴ (CP)2 = r2 ∴ (a-2)2 + ( 0 - 3)2 = 52
a2 - 4a + 4 + 9 -25 = 0
a2 - 4a -12 = 0
(a - 6) ( a + 2) = 0
a = 6 or a = -2
If a = 6 → c(6, 0), r = 5
The equation of circle will be
(x-6)2 + (y - 0)2 = 25
x2 + y2 -12x + 11 = 0
if a = -2, c = (a, 0) = (-2, 0), r = 5
The equation of circle will be
(x+2)2 + (y-0)2 = 52
x2 + y2 + 4x - 21 = 0

Question (13)

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

Solution

circle passing through (0, 0) and making intercepts a and b
So coordinate of A(a, 0) and B(0, b) Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0
It passes through (0, 0)
0 + 0 + 0 + 0 + c = 0 → c = 0
It passes through A(a, 0)
∴ a2 + 0 +2ga + 0 + c = 0
a2 + 2ga + 0 = 0 (Because c = 0)
g = -a/2
It passes through (0, b)
0 + b2 + 0 + 2 +b + 0 = 0
f = -b/2
Replacing value of g, f and c we get
x2 + y2 - ax - by = 0

Question (14)

Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Solution

The centre of the circle is given as (h, k) = (2, 2).
Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).
$r = \sqrt {{{\left( {2 - 4} \right)}^2} + {{\left( {2 - 5} \right)}^2}}$
$r = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 3} \right)}^2}} = \sqrt {4 + 9} = \sqrt {13}$
Thus, the equation of the circle is
(x - h)2 + (y - k)2 = r2
(x - 2)2 + (y - 2)2 = (√13)2
x2 - 4x + 4 + y2 - 4y + 4 = 13
x2 + y2 -4x - 4y -5 = 0

Question (15)

Does the point (–2.5, 3.5) lie inside, outside or on the circle x2 + y2 = 25?

Solution

The equation of the given circle is x2 + y2 = 25. x2 + y2 = 25 ⇒ (x – 0)2 + (y – 0)2 = 52, which is of the form (x – h)2 + (y – k)2 = r2, where h = 0, k = 0, and r = 5. ∴Centre = (0, 0) and radius = 5 Distance between point (–2.5, 3.5) and centre (0, 0) $= \sqrt {{{\left( { - 2.5 - 0} \right)}^2} + {{\left( {3.5 - 0} \right)}^2}}$
$= \sqrt {6.25 + 12.25}$
$= \sqrt {18.5}$
$= 4.3\left( {approx.} \right) < 5$
Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.