Question

A circle with its center at 0 (2,8) has a radius of 17 units.

Answer 1

Thcorrdinates of the center of the circle, (h,k)=(x,8).

The radius of the circle, r=17.

The coordinates of a point on the circle, (xo,yo)=(3,-7).

The general equation of a circle is,

$(x_0-h)^2+(y_0-k)^2=r^2$

Now, put the values in the above equation to find x.

$\begin{gathered} (3-x)^2+(-7-8)^2=17^2 \\ (3-x)^2+15^2=289 \\ (3-x)^2+225=289 \\ (3-x)^2=289-225 \\ (3-x)^2=64 \\ 3-x=\sqrt[]{64} \\ 3-x=\pm8 \end{gathered}$

Hence,

$\begin{gathered} 3-x=8\text{ } \\ 3-8=x \\ -5=x \\ or \\ 3-x=-8 \\ 3+8=x \\ 11=x \end{gathered}$

Hence, x=-5 or 11. Since it is given that x>0, the value of x is 11.