Question

Equation of the tangents drawn to the
circle $x^2+y^2-4x-8y-5=0$ which are
parallel to x-axis are:

• A. $y=\pm 5$
• B. $y=\pm 4$
• C. $y=9or-1$
• D. $y=-9or1$

Answer 1

The correct option is C $y=9or-1$

Given,

equation of the circle is

$x^2+y^2-4x-8y-5=0$

which is of the from $x^2+y^2+2yx+2fy+x=0$

$\therefore g=-2,f=-4,c=-5$

centre of the circle $-(-g,-f)=(2,4)$

radius of the circle $=\sqrt{g^2+f^2-c}=\sqrt{8^2+4^2+5}=\sqrt{4+16+5}$

$=\sqrt{15}$

$=5$

Let the equation of the length drawn to the circle with centre $(2,4)$ and radius $5$, which is parallel to the $x$ axis

i.e, $y=0$ be $y+k=0$

since $y+c=0$ is a tangent to the each,

Perpendicular distance from centre of the circle to the line

$y+k=0$ is equl to the radius.

i.e., $\frac{|0\left(2\right)+1\left(4\right)+k|}{\sqrt{0^2+1^2}}=5$

$\Rightarrow |4+k|=5$

$\Rightarrow -(4+k)=5$ or $4+k=5$

$\Rightarrow 4+k=-5k=5-4$

$\Rightarrow k=--5-4k=1$

$\Rightarrow k=-9$

Hence the equation of the tangent are

$y-9=0ory+1=0$

$\Rightarrow y=9ory-1$