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  • RBSE Class 11

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

June 15, 2019 by Fazal Leave a Comment

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

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

Question 6.
Find the centre and radius of the circle
2x2 + 2y2 – x = 0.
Solution:
According to question, equation of circle
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.1
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.1

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

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

RBSE Solutions for Class 11 Maths

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