Test Index

Circles Part 1

Section: Mathematics

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Question : 69 of 100

Marks: +1, -0

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

2x+y=3

[Main 10 Apr 2016]

$4x+y-3=0$
$x+4y+3=0$
$3x-y-4=0$
$x-3y-4=0$

Solution:

Center of circle is

\left(\frac{4}{3}, \frac{1}{3}\right)

⇒ equation of circle is

\left(x-\frac{4}{3}\right)^2+\left(y-\frac{1}{3}\right)^3=\left(1-\frac{4}{3}\right)^2+\left(-1-\frac{1}{3}\right)^2

\Rightarrow x^2-\frac{8}{3}x+\frac{16}{9}+y^2-\frac{2}{3}y+\frac{1}{9}=\frac{1}{9}+\frac{16}{9}

\Rightarrow x^2+y^2-\frac{8}{3}x-\frac{2}{3}y=0

\Rightarrow 3x^2+3y^2-8x-2y=0

Equation of tangent at

(1, -1)

is

3x-3y-4(x+1)-(y-1)=0

\Rightarrow -x-4y-3=0

\Rightarrow x+4y+3=0

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