Given two circles $x^2+y^2+8 x-6 y-24=0$ and $x^2+y^2-4 x+10 y+20=0$. Then they are
Given two circles $x^2+y^2+8 x-6 y-24=0$ and $x^2+y^2-4 x+10 y+20=0$. Then they are
- Disjoint.
- Concentric,
- Touching internally
- Touching externally.
Solution
$\begin{aligned}
& x^2+y^2+8 x-6 y-24=0, C_1 \equiv(-4,3), r_1=7 \\
& x^2+y^2-4 x+10 y+20=0, C_2 \equiv(2,-5), r_2=3 \\
& C_1 C_2=\sqrt{(2+4)^2+(-5-3)^2}=10 \text { and } r_1+r_2=7+3=10
\end{aligned}$
$\because C_1 C_2=r_1+r_2$ hence, the two circles touching externally
Asked in: MHT CET 2022 (11 Aug Shift 1)
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