In solving this problem, I first solved the slope of the perpendicular bisector and I solved the midpoint of the two points. I don't know what to do next. I hope you can help me.
$\endgroup$ 32 Answers
$\begingroup$hint...let the circle centre be $(a,b)$
You require the line from the centre to the midpoint of the chord to be perpendicular to the chord, i.e $$\frac{b-1}{a-3}=-\frac14$$and you need the distance from the centre to one of the points given to be equal to the radius, so$$(a-5)^2+(b-9)^2=85$$
Solve simultaneously.
$\endgroup$ $\begingroup$Let center to be $(a,b)$. $$(a-5)^2+(b-9)^2=85, (a-1)^2+(b+7)^2=85$$$$a^2-10a+b^2-18b=-21,a^2-2a+b^2+14b=35$$Subtracting gives $$8a+32b=56\Rightarrow a+4b=7\Rightarrow a=7-4b$$$$35=a^2-2a+b^2+14b=49-56b+16b^2-14+8b+b^2+14b=35+17b^2-34b\Rightarrow b^2-2b=0$$$$b=0\Rightarrow a=7-4b=7; b=2 \Rightarrow a=7-4b=-1$$$$(x-7)^2+y^2=85 \mathrm{\ or\ } (x+1)^2+(y-2)^2=85$$
$\endgroup$
