Question

Find the locus of mid-point of chords to the circle $(x-3{)}^2+(y-2{)}^2=1$ which pass through $(3,7)$

Answer 1

Let $P(x,y)$ be one of the points on locus,

$B(3,2)$ be the center of the circle,

$A(3,7)$

Since $AM\perp BM$, we have:

${(x-\frac{3+3}{2})}^2+(y-\frac{7+2}{2})={\left(\frac{AB}{2}\right)}^2$

$(x-2{)}^2+(y-4.5{)}^2={\left(\frac{\sqrt{(0-5^2)}}{2}\right)}^2$

$(x-2{)}^2+(y-4.5{)}^2=6.25$