Find the equation of the circle passing through two points (4,4) and (-2.-2) with radius 4

Question

I am trying to solve this question but i cannot seem to find the center (h,k). I know i have to use the equation of circle form '(x-h)^2+(x-k)^2=r^2' and after that form to solution of the 2 point given and simultaneous equations them. I got the answer for that which is h+k=2 and that is where i got stuck. Please anyone help me with this. i also post my working as well. Working

Answer 1

Since the distance between $(4,4)$ and $(-2,-2)$ is equal to $6\sqrt2$, which is greater that $8$, the problem has no solution.

Answer 2

Hint: Let $$M(x_M,y_M)$$ the coordinates of the miodpoint of the circle, then solve $$(4-x_M)^2+(4-y_M)^2=16$$ $$(-2-x_M)^2+(-2-y_M)^2=16$$ for $$x_M,y_M$$ Subtracting both equation we get $$-x_M-y_M+2=0$$ and now you can calculate $$x_M,y_M$$ substituing in one equation from above.

Answer 3

Perhaps what you are missing is that there are two circles satisfying this and two possible centers, one on each side of the line through (4, 4) and (-2, -2).

The perpendicular bisector of a secant line of a circle always passes through the center of the circle. (4, 4) and (-2, -2) are points on the circle and the line passing through both is y= x. The midpoint is (1, 1) and the perpendicular has slope -1 so the perpendicular bisector is y- 1= -(x- 1) or y= -x+ 2.

We want a point on that line such that its distance from both (4, 4) and (-2, -2) is 4. That is, we want $\sqrt{(x- 4)^2+ (y- 4)^2}= \sqrt{(x- 4)^2+ (-x+ 2- 4)^2}= \sqrt{(x- 4)^2+ (-x+ 2)^2}= 4$. Then $(x- 4)^2+ (-x+ 2)^2= x^2- 8x+ 16+ x^2- 4x+ 4= 2x^2- 12x+ 20= 16$. $x^2- 6x+ 2= x^2- 6x+ 9- 9+ 2= (x- 3)^2- 7= 0$. $x- 3= \pm\sqrt{7}$ so $x= 3\pm{7}$. If $x= 3\pm\sqrt{7}$ then $y= -(3\pm\sqrt{7})+ 2= -1\pm\sqrt{7}$. The two possible centers of the circle are $(3+ \sqrt{7}, -1- \sqrt{7})$ and $(3- \sqrt{7}, -1+ \sqrt{7})$