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

Question 1.

Find the point of intersection of circle x^{2} + y^{2} = 25 and line 4x + 3y = 12 and also find length of intersecting chord.

Solution:

Equation of circle,

x^{2} + y^{2} = 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 x^{2} +y^{2} = a^{2} cuts an intercept of length 2l at straight line mx + c, then prove that

c^{2} = (1 + m^{2})(a^{2} – l^{2}).

Solution:

Equation of circle,

x^{2} + y^{2}. = a^{2}

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 x^{2} + y^{2} = c^{2} at line \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1.

Solution:

Equation of circle,

x^{2} + y^{2} = c^{2
}

Question 4.

For what value of A, line 3x + 4y = k touches the circle x^{2} + y^{2} = 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} = r^{2}

(ii) Line lx + my + n = 0 touches the circle

x^{2} + y^{2} = a^{2}

Solution :

(i) Equation of line

y = mx + c …(i)

Equation of circle (x – a)^{2} + (y – b)^{2} = r^{2}

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 + m^{2})r^{2} = m^{2}a^{2} + b^{2} + c^{2} – 2mab – 2bc + 2mac

⇒ r^{2} + m^{2}r^{2} = m^{2}a^{2} + b^{2} + c^{2} – 2mab – 2bc + 2mac

⇒ m^{2} – a^{2} – m^{2}r^{2} + 2mac – 2mab + b^{2} + c^{2} – 2bc = r^{2}

⇒ (a^{2} – r^{2})m^{2} + (2ac – 2ab)m + b^{2} + c^{2} – 2bc = r^{2}

⇒ (a^{2} – r^{2})m^{2} + 2a(c – b)m + (b – c)^{2} = r^{2}

⇒ m^{2}(a^{2} – r^{2}) + 2ma(c – b) + (c – b)^{2} = r^{2}

(ii) Equation of given line

lx + my + n = 0

⇒ my = – lx – n

Question 6.

(i) Find the equation of tangent of the circle x^{2} + y^{2} = 64 which passes through point (4, 7).

(ii) Find the equation of tangent of the circle x^{2} + y^{2} – 4 which makes an angle of 60° with x-axis.

Solution:

(i) Equation of circle

x^{2} – y^{2} = 64

⇒ x^{2} + y^{2} = (8)^{2}

Equation of tangent passing through (4, 7)

(y – y_{1}) = m(x – x_{1})

⇒ (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 x^{2} + y^{2} – 2x + 2y – 2 = 0 at point (1, 1).

Solution:

Equation of circle,

x^{2} + y^{2} – 2x + 2y – 2 = 0 …(i)

General equation of circle,

x^{2} + y^{2} + 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 x^{2} + y^{7} = 169. Prove that they will be perpendicular to each other. Also find the coordinates of intersection point.

Solution:

Equation of circle

x^{2} + y^{2} = 169 …(i)

At point (x_{1}.y_{1}) equation of tangent line passing through the circle x^{2} + y^{2} = a^{2}

xx_{1} + yy_{1} = a^{2}

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).

Eq^{n}. (ii) × 5 and eq^{n}. (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 eq^{n}. (ii),

5 × 17 + 12y – 169 = 0

85 – 169 + 12y = 0

-84 + 12y = 0

⇒ y = \(\frac { 84 }{ 12 } \) = 7

Thus, intersection point is (17, 7).

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