Question

Pleaseee helppp with thissss asapppp

Answer 1

As we know that the standard equation of circle is
`{\bf{(x-h)^(2)+(y-k)^(2)=r^(2)}}` , 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

`{:\implies \quad \sf \{2-(-1)\}^(2)+(9-5)^(2)=r^(2)}`

`{:\implies \quad \sf (3)^(2)+(4)^(2)=r^(2)}`

`{:\implies \quad \sf 9+16=r^(2)}`

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 ;

`{:\implies \quad \sf \{x-(-1)\}^(2)+(y-5)^(2)=(5)^(2)}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{(x+1)^(2)+(y-5)^(2)=25}}}`

This is the required equation of Circle

Refer to the attachment as well !