Find the equation of the circle with radius 5 | vedclass

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Question

(B) Let the equation of the circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.Since the radius $r=5$ and the centre lies on the $x-$axis,$k=0$. Thus,the equation

Vedclass · Accepted Answer

$x^{2}+y^{2}-12x+11=0$

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(B) Let the equation of the circle be $(x-h)^{2}+(y-k)^{2}=r^{2}$.Since the radius $r=5$ and the centre lies on the $x-$axis,$k=0$. Thus,the equation is $(x-h)^{2}+y^{2}=25$.Since the circle passes through $(2, 3)$,we have $(2-h)^{2}+3^{2}=25$.$(2-h)^{2}+9=25 \Rightarrow (2-h)^{2}=16$.$2-h = \pm 4$.If $2-h=4$,then $h=-2$. The equation is $(x+2)^{2}+y^{2}=25 \Rightarrow x^{2}+y^{2}+4x-21=0$.If $2-h=-4$,then $h=6$. The equation is $(x-6)^{2}+y^{2}=25 \Rightarrow x^{2}+y^{2}-12x+11=0$.

Find the equation of the circle with radius $5$ whose centre lies on the $x-$axis and passes through the point $(2, 3)$.

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