The number of common tangents to the circles $x^2+y^2-4 x-2 y+k=0$ and $x^2+y^2-6 x-4 y+l=0$, having radii 2…
The number of common tangents to the circles $x^2+y^2-4 x-2 y+k=0$ and $x^2+y^2-6 x-4 y+l=0$, having radii 2 and 3 respectively, is
- 4
- 2
- 3
- 1
Solution
Given circle are,
$
\begin{gathered}
x^2+y^2-4 x-2 y+k=0 \\
\therefore \quad \text { Centre } c_1(2,1) \text { and } r_1=2 \\
\quad x^2+y^2-6 x-4 y+l=0
\end{gathered}
$
Centre $c_2(3,2)$ and $r_2=3$
$
\begin{aligned}
c_1 c_2 & =\sqrt{(2-3)^2+(1-2)^2} \\
& =\sqrt{1+1}=\sqrt{2} \\
c_1 c_2 & < r_1+r_2
\end{aligned}
$
So, circles have two common tangents
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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