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

Explanation

Let the centre of circle on x-axis be (h,0).
Given that radius of circle = 5
Therefore, equation of circle is
`(x-h)^(2)+(y-0)^(2)=5^(2)`
`rArr""(x-h)^(2)+y^(2)=25` . . .(1)
This circle passes through the point (2,3).
`:." "(2-h)^(2)+3^(3)=25`
`rArr" "(2-h)^(2)=16`
`rArr" "2-h=pm4`
`rArr" "h=6orh=-2`.
`:.` From equation (1)
Equation of cicle
`(x-6)^(2)+y^(2)=25or(x+2)^(2)+y^(2)=25`