If the circles $x^2+y^2-6 x-8 y-12=0$ and $x^2+y^2-4 x+6 y+k=0$ are perpendicular to each other, then ' $k$…
If the circles $x^2+y^2-6 x-8 y-12=0$ and $x^2+y^2-4 x+6 y+k=0$ are perpendicular to each other, then ' $k$ ' equals
- 4
- 0
- -2
- 12
Solution
$
\begin{aligned}
& C_1: x^2+y^2-6 x-8 y-12=0 \\
& C_2: x^2+y^2-4 x+6 y+k=0
\end{aligned}
$
Both the circles are perpendicular to each other.
$
\begin{aligned}
\because \quad g_1 & =-3, f_1=-4, c_1=-12 \\
& g_2=-2, f_2=3, c_2=k
\end{aligned}
$
$\because$ If two circles cuts perpendicularly to each other.
$
\begin{array}{rlrl}
\Rightarrow & 2 g_1 g_2+2 f_1 f_2 & =c_1+c_2 \\
\Rightarrow & 2(-3)(-2)+2(-4)(3) & =-12+k \\
\Rightarrow & 12+(-24) & =-12+k \\
\Rightarrow & & -12 & =-12+k \\
\Rightarrow & k & =0
\end{array}
$
Asked in: AP EAMCET 2021 (24 Aug Shift 1)
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