RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.1

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 = r²
[x – (- 2)]² + (y – 3)² = 42
⇒ (x + 2)² + (y – 3)² = 16
⇒ x² + 4x + 4 + y² – 6y + 9 = 16
⇒ x² + y² + 4x – 6y = 16 – 9 – 4
⇒ x² + y² + 4x – 6y = 3
⇒ x² + y² + 4x – 6y – 3 = 0
Thus, required equation of circle
x² + y² + 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 = r²
⇒ (x – a)² + (y – b)² = (a – b)²
⇒ x² + a² – 2ax + y² + b² – 2 by = a² + b² – 2ab
⇒ x² + y² – 2ax – 2by + a² + b² – a² – b² + 2ab = 0
⇒ x² + y² – 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) \(\sqrt { 1+{ k }^{ 2 } } \)(x² + y²) = 2ax + 2aky
(iii) 4(x² + y²) = 1
Solution :
(i) x(x + y – 6) = y(x – y + 8)
⇒ x² + xy – 6x = xy – y² + 8y
⇒ x² + y² + xy – 6x – xy – 8y
⇒ x² + y² – 6x – 8y = 0 …(i)
General equation of circle,
ax² + by² + 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 = a² …(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,
CA² = AD² + CD²
= 3² + 3² = 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
⇒ x² + 9 – 6x + y² + 18 – 6\(\sqrt { 2 }\) y = 18
⇒ x² + y² – 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\(\sqrt { 2 }\))
x² + y² + 6x – 6\(\sqrt { 2 }\)y + 9 = 0
x² + y² + 6x + 6\(\sqrt { 2 }\)y + 9 = 0
and x² + y² – 6x + 6\(\sqrt { 2 }\) + 9 = 0
Thus, total four circles are possible whose equations is
x² + y² ± 6x ± 6\(\sqrt { 2 }\)y + 9 = 0

Question 5.
Find the centre and radius of circle
x² + y² – 8x + 10y – 12 = 0.
Solution:
According to question, equation of circle
x² + y² – 8x + 10y – 12 = 0
⇒ x² – 8x + y² + 10y – 12 = 0
⇒ x² – 8x + 16 – 16 + y²
+ 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
2x² + 2y² – 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 = r² …(i)
According to question, circle passes through the point (2, 3) and (-1,1)
Thus (2 – h)² + (3 – k)² = r²
⇒ 4 + h² – 4h + 9 + k² – 6k = r²
and (- 1 -h)² + (1 – k)² = r²
⇒ 1 + h² + 2h + 1 + k² – 2k = r²
Again h² + k² – 4h – 6k + 13 = r² …(ii)
and h² + k² + 2h – 2k + 2 – r² …(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 + y² = 25
But circle passes through point (2, 3).
Then (2 – h)² + 3² = 25
⇒ 4 + h² – 4h + 9 = 25
⇒ h² -4h + 13 – 25 = 0
⇒ h² – 4h – 12 = 0
⇒ h² – (6 – 2)h – 12 = 0
⇒ h² – 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 + y² = 25
or x² + 36 – 12x + y² = 25
or x² + y² – 12x + 36 – 25 = 0
or x² + y² – 12x + 11 = 0
Again, putting h = – 2 in eqn. (i), equation of circle
[x – (- 2)²]² + y² = 25
⇒ (x + 2)² + y² = 25
⇒ x² + 4x + 4 + y² = 25
⇒ x² + y² + 4x + 4 – 25 = 0
⇒ x² + y² + 4x – 21 = 0
Thus, required equation of circle
X² + y² – 12x + 11 = 0
⇒ x² + y² + 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 = r² …(1)
Circle passes through (0, 0), (a, 0) and (0, b).
Thus, (0 – h)² + (0 – k)² = r² …(i)
(a – h)² + (0 – k)² = r² …(ii)
(0 – h)² + (b – k)² = r² …(iii)
h² + k² = r² …(2)
From eqⁿ. (i),
From eqⁿ. (ii),
a² + h² – 2 ah + k² = r²
⇒ h² + k² – 2 ah + a² = r² …(3)
From eqⁿ. (iii)
h² + b² + k² – 2bk = r²
⇒ h² + k² – 2bk + b² = r² …(4)
From eq11. (2) and (3)
r² – 2ah + a² = r²
⇒ -2 ah + a² – r² – r²
⇒ a² – 2h = 0
⇒ a(a -2h) = 0
Then a ≠ 0 and a – 2h = 0
⇒ h = \(\frac { a }{ 2 } \)
Similarly, from eqⁿ. (2) and (4)
r² – 2bk + b² = r²
⇒ – 2bk + b² = r² – r²
⇒ b² – 2bk = 0
⇒ b(b – 2k) = 0

RBSE Solutions for Class 11 Maths