$x^2+y^2+2 x+4 y-20=0$ and $x^2+y^2+6 x-8 y+10=0$ are the given circles. Which one of the following is…
$x^2+y^2+2 x+4 y-20=0$ and $x^2+y^2+6 x-8 y+10=0$ are the given circles. Which one of the following is correct?
They intersect orthogonally and will have two common tangents. The length of their common chord is $\frac{5 \sqrt{3}}{\sqrt{2}}$
They intersect at right angles and will have two common tangents. The length of their common chord is 2
They do not intersect orthogonally and will have three common tangents. The length of their direct common tangent is 5
They touch each other internally and will have only one common tangent
Solution
The equation of given two circles are
$
x^2+y^2+2 x+4 y-20=0
$
and
$
x^2+y^2+6 x-8 y+10=0
$
$
\because \quad \begin{aligned}
2 g_1 g_2+2 f_1 f_2 & =6-16 \\
& =-10=c_1+c_2
\end{aligned}
$
So, circles intersects each other orthogonally and will have two common tangents.
Now, equation of common chord is
$
2 x-6 y+15=0
$
So, length of common chord is
$
\begin{aligned}
& 2 \sqrt{25-\frac{(-2+12+15)^2}{4+36}} \\
= & 2 \sqrt{25-\frac{625}{40}}=2 \sqrt{\frac{1000-625}{40}}=5 \frac{\sqrt{3}}{\sqrt{2}} .
\end{aligned}
$