Aspire Faculty ID #18305 · Topic: AMU MCA 2017 · Just now
AMU MCA 2017

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

Solution

Given circle: $ x^2+y^2-2x-4y-20=0 $ $ \Rightarrow (x-1)^2+(y-2)^2=25 $ Centre $=(1,2)$, radius $=5$ New circle also radius $=5$ Since they touch externally, centres lie on line joining and midpoint is $(5,5)$ $ \Rightarrow \frac{h+1}{2}=5,\ \frac{k+2}{2}=5 $ $ \Rightarrow h=9,\ k=8 $ Equation: $ (x-9)^2+(y-8)^2=25 $

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