Question

Let the equation of the required circle be ( (x-h)^{2}+(y-k)^{2}=r^{2} )
since the radius of the circle is 5 and its centre lies on the ( x ) -axis, ( k=0 ) and ( r=5 )
Now, the equation of the circle becomes ( (x-h)^{2}+y^{2}=25 )
It is given that the circle passes through point (2,3) .
( therefore(2-h)^{2}+3^{2}=25 )
( Rightarrow(2-h)^{2}=25-9 )
( Rightarrow(2-h)^{2}=16 )
( Rightarrow 2-h=pm sqrt{16}=pm 4 )
If ( 2-h=4, ) then ( h=-2 )
If ( 2-h=-4, ) then ( h=6 )
When ( h=-2 ), the equation of the circle becomes
( (x+2)^{2}+y^{2}=25 )
( x^{2}+4 x+4+y^{2}=25 )
( x^{2}+y^{2}+4 x-21=0 )
When ( h=6, ) the equation of the circle becomes
( (x-6)^{2}+y^{2}=25 )
( x^{2}-12 x+36+y^{2}=25 )
( x^{2}+y^{2}-12 x+11=0 )

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

Solution