Question

Find the equation of the circle passing through the points $(5,5),(3,7)$ and has its center on the line $x-4y+11=0$

Answer 1

equation of circle (?)

A-(5, 5), B(3,7)

center of circle is A,B

$\to$ the center of circle has to lie on the perpendicular

bisector of line AB.

$\to$ slop of AB = $\frac{7-5}{3-5}=-1$

$\to$ Also the perpendicular bisector of AB has

to pass through mid point of AB.

$=\left(\frac{3+5}{2}\right)\left(\frac{5+7}{2}\right)$

$=(\frac{8}{2},\frac{12}{2})$

$=(4,6)$

$\to$ so, equation of perpendicular bisector

$\frac{y-6}{x-4}=I$

$y-6=x-4$

$\therefore y=x+2$

$\to$ so, now we know the center of the circle passes

through two line.

$x-4y=1$ & $y=x+2$

$\to$ solving equation

$x=-3,y=-1$

$\to$ the circle pass through (5, 5) so the radius

$=\sqrt{5-(-3{)}^2)+(5-(-1){)}^2}=\sqrt{64+36}=\sqrt{100}=10$

$\to$ so equation of circle

$(x-(-3{)}^2)+(y-(-1){)}^2=r^2$

$\therefore (x+3{)}^2+(y+1{)}^2=100$