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Question: A line parallel to the line x – 3y = 2 touches the circle x<sup>2</sup> + y<sup>2</sup> – 4x + 2y –...

A line parallel to the line x – 3y = 2 touches the circle

x2 + y2 – 4x + 2y – 5 = 0 at the point-

A

(1, –4)

B

(1, 2)

C

(3, –4)

D

(3, 2)

Answer

(1, 2)

Explanation

Solution

Let x – 3y + l = 0 touch the circle at (x1, y1).

Then xx1 + yy1 – 2(x + x1) + y + y1 – 5 = 0 and x – 3y + l = 0 are identical. Hence, comparing these

x1−21=y1+1−3=−2x1+y1−5λ\frac{x_{1} - 2}{1} = \frac{y_{1} + 1}{- 3} = \frac{- 2x_{1} + y_{1} - 5}{\lambda}

= 2(x1−2)−1(y1+1)+(−2x1+y1−5)2×1–1×(–3)+λ\frac{2(x_{1} - 2) - 1(y_{1} + 1) + ( - 2x_{1} + y_{1} - 5)}{2 \times 1–1 \times (–3) + \lambda}

\ x1−21\frac{x_{1} - 2}{1} = y1+1−3\frac{y_{1} + 1}{- 3} = −105+λ\frac{- 10}{5 + \lambda}

Ž x1 = −105+λ\frac{- 10}{5 + \lambda}+ 2 = 2λ5+λ\frac{2\lambda}{5 + \lambda}, y1 = 305+λ\frac{30}{5 + \lambda} – 1 = 25−λ5+λ\frac{25 - \lambda}{5 + \lambda}.

(x1, y1) is on the circle. So,

(2λ5+λ)2\left( \frac{2\lambda}{5 + \lambda} \right)^{2}+ (25−λ5+λ)2\left( \frac{25 - \lambda}{5 + \lambda} \right)^{2}– 4 . 2λ5+λ\frac{2\lambda}{5 + \lambda}+ 2 . 25−λ5+λ\frac{25 - \lambda}{5 + \lambda}–5 = 0.

on simplification, l2 + 10l – 75 = 0 Ž l = 5, –15.

\ x1 = 1010\frac{10}{10} = 1, y1 = 2010\frac{20}{10} = 2

or x1 = −30−10\frac{- 30}{- 10} = 3, y1 = 25+155−15\frac{25 + 15}{5 - 15} = –4.

So, (x1, y1) = (1, 2), (3, –4).