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10-1.Circle and System of Circles
hard
The common tangent to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 + 6x + 8y - 24 = 0$ also passes through the point
A
$(-4, 6)$
B
$(6, -2)$
C
$(-6, 4)$
D
$(4, -2)$
(JEE MAIN-2019)
Solution
Circle ${x^2} + {y^2} = 4$
$ \Rightarrow {c_1}\left( {0,0} \right);{r_1} = 2$
and circle ${x^2} + {y^2} + 6x + 8y – 24 = 0$
$ \Rightarrow {c_2}\left( { – 3,4} \right);{r_2} = 7$
$ \Rightarrow d = {c_1}{c_2} = 5$
also $d = \left| {{r_1} – {r_2}} \right|$
circles touch internally
Equation of common tangent ${S_1} – {S_2} = 0$
$ \Rightarrow 6x + 8y – 20 = 0$
$ \Rightarrow 3x + 4y – 10 = 0$
Point $\left( {6, – 2} \right)$ satisfy it.
Standard 11
Mathematics
