2022/057) Given $(a-5)^2 + (b-7)^2 = 4$ find the minimum of $(a+7)^2 + (b+2)^2$

$(a-5)^2 + (b-7)^2 = 4$ is a circle whose centre is $(5,7)$ and radius is 2

We need to find minimum of $(a+7)^2 + (b+2)^2$ that is distance of the point A with coordinates a,b  fom (-7,-2).

This is minimum when the point lies on the straight line from (-7,2) , (5,7) . 

this distane from (-7,2) to (5,7) = $\sqrt{(5+7)^2 + (7-2)^2} = 13$

the distance of a from (5,7) is 2

so distance of point from (-7,2)is 13 - 2 = 11

so minmum of  $(a+7)^2 + (b+2)^2$ is 121