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(D) The given circle is $x^2+y^2-2x-4y-20=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=-2$. The centre of t

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$x^2+y^2-18x-16y+120=0$

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(D) The given circle is $x^2+y^2-2x-4y-20=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $g=-1$ and $f=-2$. The centre of the given circle is $(-g, -f) = (1, 2)$. Let the centre of the required circle be $(h, k)$ and its radius be $r=5$. Since the circles touch at $(5, 5)$,the point $(5, 5)$ is the midpoint of the line segment joining the centres $(h, k)$ and $(1, 2)$. Therefore,$\frac{h+1}{2} = 5$ $\Rightarrow h+1 = 10$ $\Rightarrow h = 9$. And $\frac{k+2}{2} = 5$ $\Rightarrow k+2 = 10$ $\Rightarrow k = 8$. The equation of the required circle with centre $(9, 8)$ and radius $5$ is: $(x-9)^2 + (y-8)^2 = 5^2$ $x^2 - 18x + 81 + y^2 - 16y + 64 = 25$ $x^2 + y^2 - 18x - 16y + 145 - 25 = 0$ $x^2 + y^2 - 18x - 16y + 120 = 0$.

The equation of a circle of radius $5$ units touching another circle $x^2+y^2-2x-4y-20=0$ at $(5,5)$ is

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