If a direct common tangent drawn to the circles $x^2+y^2$ $6 x+4 y+9=0$ and $x^2+y^2+2 x-2 y+1=0$ touches…
If a direct common tangent drawn to the circles $x^2+y^2$ $6 x+4 y+9=0$ and $x^2+y^2+2 x-2 y+1=0$ touches the circles at A and B , then $\mathrm{AB}=$
- 9
- 16
- $4 \sqrt{6}$
- $2 \sqrt{6}$
Solution
$\begin{aligned}
& \text { } x^2+y^2-6 x+4 y+9=0 \\
& \Rightarrow(x-3)^2+(y+4)^2=4 ...(i)\\
& \text { and } x^2+y^2+2 x-2 y+1=0 \\
& \Rightarrow(x+1)^2+(y-1)^2=1 ...(ii)\\
& C_1=(3,-4), C_2=(-1,1), r_1=2, r_2=1
\end{aligned}$
So, $\mathrm{C}_1 \mathrm{C}_2=5\gt\left|r_1-r_2\right|=1$
$\Rightarrow$ Length of direct common tangent
$=\sqrt{\left(\mathrm{C}_1 \mathrm{C}_2\right)^2-\left(r_1-r_2\right)^2} \Rightarrow \mathrm{AB}=2 \sqrt{6}$
Asked in: AP EAMCET 2024 (21 May Shift 1)
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