11th NCERT/CBSE Introduction to Introduction to Conic section Exercise 11.1 Questions 15

Hi

Question (1)

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

The equation of a circle with centre (h, k) and radius r is given as (x . h)Question (2)

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

The equation of a circle with centre (h, k) and radius r is given as (x. h)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)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 ${\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{4}} \right)^2} = {\left( {\frac{1}{{12}}} \right)^2}$

${x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} = \frac{1}{{144}}$

${x^2} - x + \frac{1}{4} + {y^2} - \frac{y}{2} + \frac{1}{{16}} - \frac{1}{{144}} = 0$

144x

144x

36x

36x

Question (4)

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

The equation of a circle with centre (h, k) and radius r is given as (x.h)It is given that centre (h, k) = (1, 1) and radius (r) = √2.

Therefore, the equation of the circle is

(x-1)

x

x

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)

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

${\left( {x + a} \right)^2} + {\left( {y + b} \right)^2} = {\left( {\sqrt {{a^2} - {b^2}} } \right)^2}$

x

x

Question (6)

Find the centre and radius of the circle (x + 5)Solution

The equation of the given circle is (x + 5)(x + 5)

⇒ {x – (–5)}

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 xSolution

Given equation for circle xComparing it to standard form x

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 xSolution

Given equation for circle xComparing it to standard form x

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 2xSolution

Given equation for circle is 2xor ${x^2} + {y^2} - \frac{1}{2}x = 0$

Comparing it to standard form x

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 x

∴ centre of Circle, c = (-g, -f)

c∈ line 4x+y=16

∴ -4g -f - 16 = 0 ---(1)

(4,1) belongs to circle

∴ 4

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

x

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 x

centre of circle c = (-g, -f)

c ∈ x - 3y-11 = 0

∴ -g+3f - 11 = 0 ---(1)

(2, 3) ∈ circle

2

4g + 6f +c +13 = 0 ---(2)

(-1, 1) ∈ circle

(-1)

-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

${x^2} + {y^2} + 2\left( {\frac{{ - 7}}{2}} \right)x + 2\left( {\frac{5}{2}} \right)y - 14 = 0$

x

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)Let p(2,3) ∈ circle

∴ cp = r

∴ (CP)

a

a

(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)

x

if a = -2, c = (a, 0) = (-2, 0), r = 5

The equation of circle will be

(x+2)

x

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 bSo coordinate of A(a, 0) and B(0, b)

Let equation of circle be x

It passes through (0, 0)

0 + 0 + 0 + 0 + c = 0 → c = 0

It passes through A(a, 0)

∴ a

a

g = -a/2

It passes through (0, b)

0 + b

f = -b/2

Replacing value of g, f and c we get

x

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)

(x - 2)

x

x

Question (15)

Does the point (–2.5, 3.5) lie inside, outside or on the circle xSolution

The equation of the given circle is x$ = \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.