The common tangent to the circles x^2 + y^2 = | vedclass

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Question

Circle ${x^2} + {y^2} = 4$

$ \Rightarrow {c_1}\left( {0,0} \right);{r_1} = 2$

and circle ${x^2} + {y^2} + 6x + 8y - 24 = 0$

$ \Righ

Vedclass · Accepted Answer

$(6, -2)$

Vedclass · Answer

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.

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

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