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RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

June 15, 2019 by Fazal Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 1.
Find the point of intersection of circle x2 + y2 = 25 and line 4x + 3y = 12 and also find length of intersecting chord.
Solution:
Equation of circle,
x2 + y2 = 25 …(i)
Equation of line,
4x + 3y = 12 …(ii)
⇒ 3y = 12 – 4x
⇒ y = 4 – \(\frac { 4 }{ 3 } \) x …(iii)
From equation (i) and (iii),
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 2.
If circle x2 +y2 = a2 cuts an intercept of length 2l at straight line mx + c, then prove that
c2 = (1 + m2)(a2 – l2).
Solution:
Equation of circle,
x2 + y2. = a2
and equation of line,
y = mx + c
Length of intercept = 2l
Formula for length of intercept,
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 3.
Find the length of intercept cut by circle x2 + y2 = c2 at line \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1.
Solution:
Equation of circle,
x2 + y2 = c2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 4.
For what value of A, line 3x + 4y = k touches the circle x2 + y2 = 10x
Solution:
Equation of line,
3x + 4y = k
⇒ 4y = – 3x + k
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 5.
Find the condition, when
(i) Line y = mx + c, touches the circle
(x – a)2 + (y – b)2 = r2
(ii) Line lx + my + n = 0 touches the circle
x2 + y2 = a2
Solution :
(i) Equation of line
y = mx + c …(i)
Equation of circle (x – a)2 + (y – b)2 = r2
Centre of circle = (a, b) and Radius = r
Line will touch (i) and (ii), if
Radius of circle = length of ± drawn from centre to line
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2
⇒ (1 + m2)r2 = m2a2 + b2 + c2 – 2mab – 2bc + 2mac
⇒ r2 + m2r2 = m2a2 + b2 + c2 – 2mab – 2bc + 2mac
⇒ m2 – a2 – m2r2 + 2mac – 2mab + b2 + c2 – 2bc = r2
⇒ (a2 – r2)m2 + (2ac – 2ab)m + b2 + c2 – 2bc = r2
⇒ (a2 – r2)m2 + 2a(c – b)m + (b – c)2 = r2
⇒ m2(a2 – r2) + 2ma(c – b) + (c – b)2 = r2
(ii) Equation of given line
lx + my + n = 0
⇒ my = – lx – n
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 6.
(i) Find the equation of tangent of the circle x2 + y2 = 64 which passes through point (4, 7).
(ii) Find the equation of tangent of the circle x2 + y2 – 4 which makes an angle of 60° with x-axis.
Solution:
(i) Equation of circle
x2 – y2 = 64
⇒ x2 + y2 = (8)2
Equation of tangent passing through (4, 7)
(y – y1) = m(x – x1)
⇒ (y – 7) = m(x – 4)
⇒ y – 7 = mx – 4m
⇒ mx – y – 4m + 7 = 0 …(i)
∵ line of Eq. (i) touches the circle
∴ Perpendicular drawn from centre (0, 0) of circle to the tangent will be equal to radius of circle,
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 7.
Find the value of c, where line y = c, touches the circle x2 + y2 – 2x + 2y – 2 = 0 at point (1, 1).
Solution:
Equation of circle,
x2 + y2 – 2x + 2y – 2 = 0 …(i)
General equation of circle,
x2 + y2 + 2gx + 2fy + c = 0 …(ii)
Comparing equation (i) and (ii),
2g = -2 ⇒ g = -1
2f = 2 ⇒ f = 1.
c = -2
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 8.
Find the equation of tangent at point (5, 12) and (12, – 5) at circle x2 + y7 = 169. Prove that they will be perpendicular to each other. Also find the coordinates of intersection point.
Solution:
Equation of circle
x2 + y2 = 169 …(i)
At point (x1.y1) equation of tangent line passing through the circle x2 + y2 = a2
xx1 + yy1 = a2
At point (5, 12)
x × 5 + y × 12 = 169
⇒ 5x + 12y – 169 = 0 …(ii)
⇒ Gradient of this line,
RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Ex 12.2
Thus, line (ii) and (iii) are ⊥r to each other.
For intersection point, solving equation (ii) and (iii).
Eqn. (ii) × 5 and eqn. (iii) × 12
(5x + 12y = 169) × 5
(12x -Sy= 169) × 12
⇒ 25x + 60y = 845
144x – 60y = 2028
169x = 2873
x = \(\frac { 2873 }{ 169 } \) = 17
Put value of x in eqn. (ii),
5 × 17 + 12y – 169 = 0
85 – 169 + 12y = 0
-84 + 12y = 0
⇒ y = \(\frac { 84 }{ 12 } \) = 7
Thus, intersection point is (17, 7).

RBSE Solutions for Class 11 Maths

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