The extremities of a diameter of a circle.. The extremities of a diameter of a circle have co ordinates $(6,-2)$ and $(6,8)$. What length does the circle intercept along $X$ axis..?
My Attempt:
Let, $(6,-2)\equiv (x_1,y_1)$ and $(6,8)\equiv (x_2,y_2)$ be the end points of diameter of circle.
Now, 
$$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$
$$(x-6)(x-6)+(y+2)(y-8)=0$$
$$x^2-12x+36+y^2-6y-16=0$$
$$x^2+y^2-12x-6y+20=0$$
is the required equation of the circle. Now, what should I do then?

Continue with your solution -
The lengths of intercepts made by the circle 
$$x^2 + y^2 + 2gx + 2fy + c = 0$$ 
with X and Y axes are $2\sqrt{g^2−c}$ and $2\sqrt{f^2−c}$ respectively.

Or -

Mid point of (6,-2) and (6,8) is (6,3).
Also radius r = $\frac 12 \sqrt{(6-6)^2 + (8+2)^2}$
= $\frac 12 \sqrt {100}$
= 5
Equation of circle is -
$(x - 6)^2 + (y - 3)^2 = (5)^2$
Put y = 0, you will get two values of x 
subtract the two intercepts to get the length of the chord.
