Equation of tangent to circle, at point (1 , -1) whose centre is point of intersection of straight l

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10-1.Circle and System of Circles

hard

Equation of the tangent to the circle, at the point $(1 , -1)$ whose centre is the point of intersection of the straight lines $x - y = 1$ and $2x + y= 3$ is

A

$x + 4y+ 3 = 0$

B

$3x - y- 4 = 0$

C

$x-3y-4 = 0$

D

$4x + y- 3 = 0$

(JEE MAIN-2016)

Solution

Point of intersection of lines

$x-y=1$ and $2x+y=3$ is $\left( {\frac{4}{3},\frac{1}{3}} \right)$

Slope of $OP = \frac{{\frac{1}{3} + 1}}{{\frac{4}{3} – 1}} = \frac{{\frac{4}{3}}}{{\frac{1}{3}}} = 4$

Slope of tangent $ = – \frac{1}{4}$

Equation of tangent $y + 1 = – \frac{1}{4}\left( {x – 1} \right)$

$4y + 4 = – x + 1$

$x + 4y + 3 = 0$

Standard 11

Mathematics

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