Find the equation of the circle whose two end points of the diameter are $(4, - 2)$ and $(- 1, 3)$ .

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Solution

Given the two end points of the diameter of the circle is $(4, - 2)$ and $(- 1, 3)$ .
Then the equation of the circle is
$(x - 4) (x + 1) + (y + 2) (y - 3) = 0$
or, $x^{2} + y^{2} - 3 x - y - 10 = 0$
or, $x^{2} + y^{2} - 3 x - y = 10$ .

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Find the equation of the circle whose two end points of the diameter are (4,−2) and (−1,3).

Given the two end points of the diameter of the circle is (4,−2) and (−1,3).Then the equation of the circle is(x−4)(x+1)+(y+2)(y−3)=0or, x2+y2−3x−y−10=0or, x2+y2−3x−y=10.

Find the equation of the circle whose two end points of the diameter are $(4, - 2)$ and $(- 1, 3)$ .

Given the two end points of the diameter of the circle is $(4, - 2)$ and $(- 1, 3)$ .
Then the equation of the circle is
$(x - 4) (x + 1) + (y + 2) (y - 3) = 0$
or, $x^{2} + y^{2} - 3 x - y - 10 = 0$
or, $x^{2} + y^{2} - 3 x - y = 10$ .