Question

R (-3,1) and S (-1,3) are points on a circle. If RS is a diameter, find the equation of the circle.​

Answer 1

Answer:

$\sf (x+2)^2+(y-2)^2=2$

Step-by-step explanation:

If RS is the diameter of the circle, then the midpoint of RS will be the center of the circle.

$\sf midpoint=\left(\dfrac{x_s-x_r}{2}+x_r,\dfrac{y_s-y_r}{2}+y_r \right)$

             $\sf =\left(\dfrac{-1-(-3)}{2}+(-3),\dfrac{3-1}{2}+1 \right)$

             $\sf =(-2, 2)$

Equation of a circle:   $\sf (x-h)^2+(y-k)^2=r^2$

(where (h, k) is the center and r is the radius)

Substituting found center (-2, 2) into the equation of a circle:

$\sf \implies (x-(-2))^2+(y-2)^2=r^2$

$\sf \implies (x+2)^2+(y-2)^2=r^2$

To find $\sf r^2$, simply substitute one of the points into the equation and solve:

$\sf \implies (-3+2)^2+(1-2)^2=r^2$

$\sf \implies 1+1=r^2$

$\sf \implies r^2=2$

Therefore, the equation of the circle is:

$\sf (x+2)^2+(y-2)^2=2$