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Question: If a circle, whose centre is (–1, 1) touches the straight line \(x + 2y + 12 = 0,\) then the coordin...

If a circle, whose centre is (–1, 1) touches the straight line x+2y+12=0,x + 2y + 12 = 0, then the coordinates of the point of contact are

A

(−72,−4)\left( - \frac{7}{2}, - 4 \right)

B

x2+y2−2x+6y+a=0,x^{2} + y^{2} - 2x + 6y + a = 0,

C

(2, –7)

D

(– 2, – 5)

Answer

x2+y2−2x+6y+a=0,x^{2} + y^{2} - 2x + 6y + a = 0,

Explanation

Solution

Let point of contact be P(x1,y1).P(x_{1},y_{1}).

This point lies on the given line ,

∴x1+2y1=−12x_{1} + 2y_{1} = - 12 ….. (i)

Gradient of OP=m1=y1−1x1+1OP = m_{1} = \frac{y_{1} - 1}{x_{1} + 1},Gradient of

x+2y+12=m2=−12x + 2y + 12 = m_{2} = - \frac{1}{2}Both are perpendicular,

∴m1m2=−1\therefore m_{1}m_{2} = - 1

⇒(y1−1x1+1)(−12)=−1\Rightarrow \left( \frac{y_{1} - 1}{x_{1} + 1} \right)\left( \frac{- 1}{2} \right) = - 1

⇒y1−1=2x1+2\Rightarrow y_{1} - 1 = 2x_{1} + 2

⇒2x1−y1=−3\Rightarrow 2x_{1} - y_{1} = - 3….. (ii)

On solving the equation (i) and (ii), (x1,y1)=(−185,−215)(x_{1},y_{1}) = \left( \frac{- 18}{5},\frac{- 21}{5} \right)