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

User Pons Purushothaman
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1 Answer

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29 votes

Answer:


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

Explanation:

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


\sf midpoint=\left((x_s-x_r)/(2)+x_r,(y_s-y_r)/(2)+y_r \right)


\sf =\left((-1-(-3))/(2)+(-3),(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

User Deloris
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