As we know that the standard equation of circle is [tex]{\bf{(x-h)^{2}+(y-k)^{2}=r^{2}}}[/tex] , where r is the radius of circle and centre at (h,k)
Now , as the circle passes through (2,9) so it must satisfy the above equation after putting the values of h and k respectively
[tex]{:\implies \quad \sf \{2-(-1)\}^{2}+(9-5)^{2}=r^{2}}[/tex]
[tex]{:\implies \quad \sf (3)^{2}+(4)^{2}=r^{2}}[/tex]
[tex]{:\implies \quad \sf 9+16=r^{2}}[/tex]
After raising ½ power to both sides , we will get r = +5 , -5 , but as radius can never be -ve . So r = +5
Now , putting values in our standard equation ;
[tex]{:\implies \quad \sf \{x-(-1)\}^{2}+(y-5)^{2}=(5)^{2}}[/tex]
[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^{2}+(y-5)^{2}=25}}}[/tex]
This is the required equation of Circle
Refer to the attachment as well !
