-2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Paul Jun 21 '16 at 12:50
  • $\begingroup$ 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 ? $\endgroup$ – Yves Daoust Jun 21 '16 at 12:50
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – C.dave Jun 21 '16 at 13:16
  • $\begingroup$ 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. $\endgroup$ – boaz Jun 22 '16 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.