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The equation of the circle whose radius is $5$ and which touches the circle ${x^2} + {y^2} - 2x - 4y - 20 = 0$ externally at the point $(5, 5)$ is
${x^2} + {y^2} - 18x - 16y - 120 = 0$
${x^2} + {y^2} - 18x - 16y + 120 = 0$
${x^2} + {y^2} + 18x + 16y - 120 = 0$
${x^2} + {y^2} + 18x - 16y + 120 = 0$
Solution
(b) Let the centre of the required circle be $({x_1},\;{y_1})$ and the centre of given circle is $(1, 2)$.
Since radii of both circles are same, therefore, point of contact $(5, 5)$ is the mid point of the line joining the centres of both circles.
Hence ${x_1} = 9$ and ${y_1} = 8$. Hence the required equation is ${(x – 9)^2} + {(y – 8)^2} = 25$
$ \Rightarrow {x^2} + {y^2} – 18x – 16y + 120 = 0$.
Trick : The point $(5, 5)$ must satisfy the required circle. Hence the required equation is given by $(b)$.
