Find the equation of the circle which touches | vedclass

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(A) Let the center of the circle be $(h, k)$. Since the circle touches the $X$-axis at $(3, 0)$,the $x$-coordinate of the center is $h = 3$ and the ra

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$x^2 + y^2 - 6x - 10y + 9 = 0$

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(A) Let the center of the circle be $(h, k)$. Since the circle touches the $X$-axis at $(3, 0)$,the $x$-coordinate of the center is $h = 3$ and the radius $r = |k|$.Thus,the equation of the circle is $(x - 3)^2 + (y - k)^2 = k^2$.This simplifies to $x^2 - 6x + 9 + y^2 - 2ky + k^2 = k^2$,or $x^2 + y^2 - 6x - 2ky + 9 = 0$.The circle cuts an intercept of $8$ on the positive $Y$-axis. Setting $x = 0$,we get $y^2 - 2ky + 9 = 0$.The length of the intercept on the $Y$-axis is given by $2\sqrt{f^2 - c}$,where $f = -k$ and $c = 9$.So,$2\sqrt{(-k)^2 - 9} = 8 \implies \sqrt{k^2 - 9} = 4$.Squaring both sides,$k^2 - 9 = 16 \implies k^2 = 25 \implies k = 5$ (since it is on the positive $Y$-axis).Substituting $k = 5$ into the equation,we get $x^2 + y^2 - 6x - 2(5)y + 9 = 0$,which is $x^2 + y^2 - 6x - 10y + 9 = 0$.

Find the equation of the circle which touches the $X$-axis at a distance of $+3$ from the origin and cuts an intercept of $8$ on the positive $Y$-axis.

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