## Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.1

Question 1.

Find the equation of the circle whose:

(i) Centre (- 2, 3) and radius is 4

(ii) Centre (a, b) and radius is a – b

Solution:

(i) According to question,

Centre of circle (h, k) = (- 2, 3)

and Radius of circle r = 4 unit

Then, equation of circle

From formula (x – h)^{2} + (y- k)^{2} = r^{2}

[x – (- 2)]^{2} + (y – 3)^{2} = 42

⇒ (x + 2)^{2} + (y – 3)^{2} = 16

⇒ x^{2} + 4x + 4 + y^{2} – 6y + 9 = 16

⇒ x^{2} + y^{2} + 4x – 6y = 16 – 9 – 4

⇒ x^{2} + y^{2} + 4x – 6y = 3

⇒ x^{2} + y^{2} + 4x – 6y – 3 = 0

Thus, required equation of circle

x^{2} + y^{2} + 4x – 6y – 3 = 0.

(ii) According to question,

Centre of circle (h, k) = (a, b)

and Radius of circle r = (a – b) unit

Then, equation of circle

From formula (x – h)^{2} + (y – k)^{2} = r^{2}

⇒ (x – a)^{2} + (y – b)^{2} = (a – b)^{2}

⇒ x^{2} + a^{2} – 2ax + y^{2} + b^{2} – 2 by = a^{2} + b^{2} – 2ab

⇒ x^{2} + y^{2} – 2ax – 2by + a^{2} + b^{2} – a^{2} – b^{2} + 2ab = 0

⇒ x^{2} + y^{2} – lax – 2by + 2ab = 0.

Question 2.

Find the coordinates of centre and radius of the following circles :

(i) x(x + y – 6) = y (x – y + 8)

(ii) \(\sqrt { 1+{ k }^{ 2 } } \)(x^{2} + y^{2}) = 2ax + 2aky

(iii) 4(x^{2} + y^{2}) = 1

Solution :

(i) x(x + y – 6) = y(x – y + 8)

⇒ x^{2} + xy – 6x = xy – y^{2} + 8y

⇒ x^{2} + y^{2} + xy – 6x – xy – 8y

⇒ x^{2} + y^{2} – 6x – 8y = 0 …(i)

General equation of circle,

ax^{2} + by^{2} + 2gx + 2fy + c = 0 …(ii)

Comparing equation (i) and (ii),

2g = -6 ⇒ g = -3

2f = -8 ⇒ f = – 4

c = 0

Thus centre of circle

Standard equation of circle,

(x – h)^{2} + (y – k)^{2} = a^{2} …(ii)

Comparing equation (i) and (ii),

h = 0, k = 0, a = \(\frac { 1 }{ 2 } \)

Thus centre of circle is (0, 0) and radius is 1/2.

Question 3.

Find the equation of a circle which touches y-axis and cuts an intercept of length 2/on x-axis.

Solution:

In the standard of equation of circle put

Question 4.

Find the equation of the circle which cuts x- axis at a distance +3 from origin and cuts on intercept at y-axis of length 6 units.

Solution:

Let circle touches x-axis at point E and cuts AB intercepts at y-axis.

According to questions,

Let C is centre of the circle. Draw CD ⊥AB. Then CD = OE = 3 and

AD = \(\frac { AB }{ 2 } \) = \(\frac { 6 }{ 2 } \) = 3

In right angled triangle ACD,

CA^{2} = AD^{2} + CD^{2}

= 3^{2} + 3^{2} = 9 + 9 = 18

= 2 × 9 = 3\(\sqrt { 2 }\)

∴ Radius of circle a = CA = 3\(\sqrt { 2 }\)

whereas CE = CA = 3\(\sqrt { 2 }\)

Thus, centre of circle will be (3, 3\(\sqrt { 2 }\))

Equation of circle,

(x – 3)^{2} – (y – 3\(\sqrt { 2 }\))^{2} = (3\(\sqrt { 2 }\))^{2}

⇒ x^{2} + 9 – 6x + y^{2} + 18 – 6\(\sqrt { 2 }\) y = 18

⇒ x^{2} + y^{2} – 6x – 6\(\sqrt { 2 }\) y + 9 = 0

Similarly circle will be in IInd, IIIrd and IVth quadrant whose centres will be

(- 3, 3\(\sqrt { 2 }\)), (- 3, – 3\(\sqrt { 2 }\)), (3, – 3\(\sqrt { 2 }\))

x^{2} + y^{2} + 6x – 6\(\sqrt { 2 }\)y + 9 = 0

x^{2} + y^{2} + 6x + 6\(\sqrt { 2 }\)y + 9 = 0

and x^{2} + y^{2} – 6x + 6\(\sqrt { 2 }\) + 9 = 0

Thus, total four circles are possible whose equations is

x^{2} + y^{2} ± 6x ± 6\(\sqrt { 2 }\)y + 9 = 0

Question 5.

Find the centre and radius of circle

x^{2} + y^{2} – 8x + 10y – 12 = 0.

Solution:

According to question, equation of circle

x^{2} + y^{2} – 8x + 10y – 12 = 0

⇒ x^{2} – 8x + y^{2} + 10y – 12 = 0

⇒ x^{2} – 8x + 16 – 16 + y^{2}

+ 10y + 25 – 25 – 12 = 0

(On completing square)

⇒ (x – 4)^{2} + (y + 5)^{2} – 53 = 0

⇒ (x – 4)^{2} + [y – (- 5)]^{2} = 53

Question 6.

Find the centre and radius of the circle

2x^{2} + 2y^{2} – x = 0.

Solution:

According to question, equation of circle

Question 7.

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

Solution:

The equation of circle

(x – h)^{2} + (y – k)^{2} = r^{2} …(i)

According to question, circle passes through the point (2, 3) and (-1,1)

Thus (2 – h)^{2} + (3 – k)^{2} = r^{2}

⇒ 4 + h^{2} – 4h + 9 + k^{2} – 6k = r^{2}

and (- 1 -h)^{2} + (1 – k)^{2} = r^{2}

⇒ 1 + h^{2} + 2h + 1 + k^{2} – 2k = r^{2}

Again h^{2} + k^{2} – 4h – 6k + 13 = r^{2} …(ii)

and h^{2} + k^{2} + 2h – 2k + 2 – r^{2} …(iii)

Since centre of circle (h, k) lie on line

x – 3y – 11 = 0

Thus h – 3k – 11 – 0

⇒ h – 3k – 11 …(iv)

Subtracting eq. (iii) from (ii),

Question 8.

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

Solution:

According to question, centre of circle lies on x-axis. Let centre of circle (h, 0) and radius = 5 unit.

Then equation of circle

(x – h)^{2} + (y – 0)^{2} = 52

⇒ (x – h)^{2} + y^{2} = 25

But circle passes through point (2, 3).

Then (2 – h)^{2} + 3^{2} = 25

⇒ 4 + h^{2} – 4h + 9 = 25

⇒ h^{2} -4h + 13 – 25 = 0

⇒ h^{2} – 4h – 12 = 0

⇒ h^{2} – (6 – 2)h – 12 = 0

⇒ h^{2} – 6h + 2h – 12 = 0

⇒ h(h – 6) + 2(h – 6) = 0

⇒ (h – 6) (h + 2) = 0

⇒ h = 6 or h = -2

Then centre of each (6, 0) or (- 2, 0)

Put h = 6 in equation (1), the equation of circle

(x – 6)^{2} + y^{2} = 25

or x^{2} + 36 – 12x + y^{2} = 25

or x^{2} + y^{2} – 12x + 36 – 25 = 0

or x^{2} + y^{2} – 12x + 11 = 0

Again, putting h = – 2 in eqn. (i), equation of circle

[x – (- 2)^{2}]^{2} + y^{2} = 25

⇒ (x + 2)^{2} + y^{2} = 25

⇒ x^{2} + 4x + 4 + y^{2} = 25

⇒ x^{2} + y^{2} + 4x + 4 – 25 = 0

⇒ x^{2} + y^{2} + 4x – 21 = 0

Thus, required equation of circle

X^{2} + y^{2} – 12x + 11 = 0

⇒ x^{2} + y^{2} + 4x – 21 = 0

Question 9.

Find the equation of circle that passes through point (0, 0) and cuts intercepts a and b at axis.

Solution:

According to question, circle passes through (0, 0) and cut intercepts a and b at x and y-axis.

Intersection points of circle with x-axis will be (a, 0) and with y-axis will be (0, b).

Thus, circle will passed through three points (0, 0), (a, 0) and (0, b).

Let equation of circle,

(x – h)^{2} + (y – k)^{2} = r^{2} …(1)

Circle passes through (0, 0), (a, 0) and (0, b).

Thus, (0 – h)^{2} + (0 – k)^{2} = r^{2} …(i)

(a – h)^{2} + (0 – k)^{2} = r^{2} …(ii)

(0 – h)^{2} + (b – k)^{2} = r^{2} …(iii)

h^{2} + k^{2} = r^{2} …(2)

From eq^{n}. (i),

From eq^{n}. (ii),

a^{2} + h^{2} – 2 ah + k^{2} = r^{2}

⇒ h^{2} + k^{2} – 2 ah + a^{2} = r^{2} …(3)

From eq^{n}. (iii)

h^{2} + b^{2} + k^{2} – 2bk = r^{2}

⇒ h^{2} + k^{2} – 2bk + b^{2} = r^{2} …(4)

From eq11. (2) and (3)

r^{2} – 2ah + a^{2} = r^{2}

⇒ -2 ah + a^{2} – r^{2} – r^{2}

⇒ a^{2} – 2h = 0

⇒ a(a -2h) = 0

Then a ≠ 0 and a – 2h = 0

⇒ h = \(\frac { a }{ 2 } \)

Similarly, from eq^{n}. (2) and (4)

r^{2} – 2bk + b^{2} = r^{2}

⇒ – 2bk + b^{2} = r^{2} – r^{2}

⇒ b^{2} – 2bk = 0

⇒ b(b – 2k) = 0

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