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

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),

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,

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

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

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

⇒ (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

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,

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

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,

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