Finding an equation of a circle

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.

• In the first problem use the two points to find the radius of the circle. Now you can write down the circle equation. In the second problem, can you identify a point on the circle? - it may help to do a little sketch. Now you can find the radius etc. – Paul Jun 21 '16 at 12:50
• Hou can you express that you know the coordinates of the center ? How can you express that you know a point on the circle ? How can you express tangency to an axis ? – Yves Daoust Jun 21 '16 at 12:50

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?

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 $$(x+4)^2+(y-6)^2=16$$

• I'm a little bit confused sir on how did you get this equation, (x−a)2+(y−b)2=(x0−a)2+(y0−b). – C.dave Jun 21 '16 at 13:16
• since $R=\sqrt{(x_0-a)^2+(y_0-b)^2}$, it follows that $R^2=(x_0-a)^2+(y_0-b)^2$. Now plug it into the circle equation. – boaz Jun 22 '16 at 5:32