How do you find general form of circle with endpoints of a diameter at (4,3) and (0,1)?

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Answer 1 · 2018-06-14T07:39:25

$x^{x} + y^{2} + 4 x + 4 y - 9 = 0$ $x^x+y^2+4x+4y-9=0$

eqn. of circle

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x-a)^2+(y-b)^2=r^2$

$(a, b) = the centre$ $(a,b)=" the centre"$

$r = the radius$ $r=" the radius"$

centre is the midpoint of the diameter

$(a, b) = (\frac{4 + 0}{2}, \frac{3 + 1}{2})$ $(a,b)=((4+0)/2,(3+1)/2)$

$(a, b) = (2, 2)$ $(a,b)=(2,2)$

$r = \sqrt{{(2 - 0)}^{2} + {(2 - 1)}^{2}}$ $r=sqrt((2-0)^2+(2-1)^2)$

$r = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$ $r=sqrt(4^2+1^2)=sqrt17$

eqn.

${(x - 2)}^{2} + {(y - 2)}^{2} = 17$ $(x-2)^2+(y-2)^2=17$

$x^{2} + y^{2} - 4 x - 4 y + 8 = 17$ $x^2+y^2-4x-4y+8=17$

$x^{x} + y^{2} - 4 x - 4 y - 9 = 0$ $x^x+y^2-4x-4y-9=0$