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 !