Mathematics - Circle Question with Solution | TestHub

Given equation of circle $(x+2{)}^2+(y-3{)}^2=4$ Let the coordinates of Q is $(h,k)$.Coordinate of B which is midpoint of AQ because $\mathrm{AQ}=2\mathrm{AB}$. Then, $\mathrm{B}=\left(\frac{0+h}{2},\frac{k+3}{2}\right)\to \left(\frac{h}{2},\frac{k+3}{2}\right)$ Point B also satisfy the equation of circle. $(x+2{)}^2+(y-3{)}^2=4$ ${\left(\frac{h}{2}+2\right)}^2+{\left(\frac{k+3}{2}-3\right)}^2=4$ $\frac{(h+4{)}^2}{4}+\frac{(k-3{)}^2}{4}=4$ $(h+4{)}^2+(k-3{)}^2=16$ Replace $(h,k)$ by $(x,y)$, then, the required equation is $(x+4{)}^2+(y-3{)}^2=16$