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A circle has a diameter with endpoints (-10, -6) and (-2, -4). What is the equation of the circle?
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(x-x1)(x-x2) + (y-y1)(y-y2) =0
where (x1,y1) and (x2,x2) are given end points of a diameter
r2 = (x + 4)2 + (y + 5)2 r2 = (x + 6)2 + (y + 5)2 r2 = (x + 4)2 + (y + 1)2 or r2 = (x + 6)2 + (y - 1)2 are my only answer options
The answers are not adequate because r^2 is not given. In a case like this, you may be better off finding the centre of the circle and use the standard equations. The centre of the circle is given by the mid-point between the two given points: \((xm,ym)=((xa+xb)/2, (ya+yb)/2)\) The equation of the circle is then \((x-xm)^2+(y-ym)^2 = r^2\) whatever r=radius may be.
ok thanks
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