Question

Write down the equation of the tangent to the circle :
$4x^2+4y^2-3x-16y+24y=117$ at the point $(-4,-\frac{11}{2})$

• A. $36y+35x+338=0$
• B. $35y+36x+338=0$
• C. $35y+36x-330=0$
• D. $36(y+x)+338=0$

Answer 1

The correct option is A $36y+35x+338=0$

The centre $O$ of the circle $4x^2+4y^2-3x+8y=117$ is $(\frac{3}{8},-1)$.

Let the point $(-4,-\frac{11}{2})$ be $P$

So, the slope of line $OP$ is

$m_1=\frac{-\frac{11}{2}+1}{-4-\frac{3}{8}}$

$\Rightarrow m_1=\frac{36}{35}$

Let the slope of tangent to circle is $m_2$

So, $m_1.m_2=-1$

So, $m_2=-\frac{35}{36}$

Equation of tangent to the circle is,

$y+\frac{11}{2}=-\frac{35}{36}(x+4)$

$\Rightarrow 36y+35x+338=0$