If the circles $x^2+y^2-8 x-8 y+28=0$ and $x^2+y^2-8 x-6 y$ $+25-\alpha^2=0$ have only one common tangent,…
If the circles $x^2+y^2-8 x-8 y+28=0$ and $x^2+y^2-8 x-6 y$ $+25-\alpha^2=0$ have only one common tangent, then $\alpha=$
- $\alpha=4$
- $\alpha=2$
- $\alpha=1$
- $\alpha=5$
Solution
$S_1=x^2+y^2-8 x-8 y+28=0$
$\Rightarrow C_1=(4,4), r_1=\sqrt{16+16-28}=2$
$S_2=x^2+y^2-8 x-6 y+25-\alpha^2=0$
$\Rightarrow C_2=(4,3), r_2=\sqrt{16+9-25+\alpha^2}=\alpha$
Condition for one common tangent is
$\left|r_1-r_2\right|=C_1 C_2 \Rightarrow(2-\alpha)^2=1 \Rightarrow \alpha=1$
Asked in: AP EAMCET 2024 (20 May Shift 1)
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