Two circles, each of radius $5$ units touch each other at $(1, 2)$ . If their common tangent is $4 x + 3 y = 10$ then the equations of the circles are

x^{2} + y^{2} + 10 x + 10 y + 25 = 0

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x^{2} + y^{2} + 6 x + 2 y - 15 = 0

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x^{2} + y^{2} - 10 x - 10 y + 25 = 0

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x^{2} + y^{2} + 6 x + 2 y + 15 = 0

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Solution

The correct options are
B $x^{2} + y^{2} + 6 x + 2 y - 15 = 0$
D $x^{2} + y^{2} - 10 x - 10 y + 25 = 0$
$4 x + 3 y = 10$ is of the form $a x + b y = c$
whose slope is $\frac{- a}{b}$
Thus, slope of $4 x + 3 y = 10$ is $\frac{- 4}{3}$
As the line $O_{1} O_{2}$ is perpendicular to the line $4 x + 3 y = 10$ we have
slope of the line $O_{1} O_{2} \times$ slope of $4 x + 3 y = 10 = - 1$
Thus, slope of $O_{1} O_{2} = \frac{- 1}{\frac{- 4}{3}} = \frac{3}{4}$
$∴ tan θ = \frac{3}{4}$
$\Rightarrow sin θ = \frac{3}{5}, cos θ = \frac{4}{5}$
The centre is at $(x, y)$ where
$x = 1 + 5 cos θ$ and $y = 2 + 5 sin θ$
$x = 5, y = 5$
The midpoint of $O_{1} O_{2}$ is $(1, 2)$
$∴ O_{1} = (- 3, - 1)$
The circles are ${(x - 5)}^{2} + {(y - 5)}^{2} = 5^{2}$
and ${(x + 3)}^{2} + {(y + 1)}^{2} = 5^{2}$
or $x^{2} + y^{2} - 10 x - 10 y + 25 = 0$ and $x^{2} + y^{2} + 6 x + 2 y - 15 = 0$

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