My math homework are finding an equation of a circle. Given that the center is at (-10,0) and passes through A(-6,3). Second item is the given center is at (-4, 6) and is tangent to the axis.
I've no idea how to solve this because the examples in our book aren't clear.
On
The equation of a circle with center $O(a,b)$ and radios $R$ is $$ (x-a)^2+(y-b)^2=R^2 $$ If $A(x_0,y_0)$ is a point on the circle, then the radios is $R=\sqrt{(x_0-a)^2+(y_0-b)^2}$, so given the center $O$ and a point $A$ on the circle, the equation is $$ (x-a)^2+(y-b)^2=(x_0-a)^2+(y_0-b)^2 $$ In your example, the equation is $$ (x+10)^2+(y-0)^2=(-6+10)^2+(3-0)^2 $$ that is $$ (x+10)^2+y^2=25 $$
for the part (b), the radios is $R=4$, since the circle is tangent to the axis (y axis in this case - if it unclear, just draw it). Hence, the equation is $$ (x+4)^2+(y-6)^2=16 $$
Hint:
I suppose that you know that the equation of a circle of radius $r$ an center in a point $C=(\alpha,\beta)$ is: $$ (x-\alpha)^2+(y-\beta)^2=r^2 $$
you r first circle has center $C=(-1,0)$ and the radius is the distance $r=\overline{CA}$.
For the second circle the radius is the distance from the given center and the tangent axis.
can you do from this?