# Math Help - Circles?

1. ## Circles?

Write the equation of the circle that satisfies each set of conditions.

1.The endpoints of the diameters are (5,6) and (2,3)?

2. The circle passes through origin and has its center at (-4,3).

Now I know the standard form of the equation of a circle is

(x-h)^2+(y-k)^2=r^2

But how would I solve these two problems I mean for number 2 I know x and y but no the radius.

2. Originally Posted by homeylova223
Write the equation of the circle that satisfies each set of conditions.

1.The endpoints of the diameters are (5,6) and (2,3)?

2. The circle passes through origin and has its center at (-4,3).

Now I know the standard form of the equation of a circle is

(x-h)^2+(y-k)^2=r^2

But how would I solve these two problems I mean for number 2 I know x and y but no the radius.
1) You have two points on the circle and you know they represent a diameter. How do you find the diameter from this? The center of the circle is at the center of the diameter. How do you find this?

2) You have the center so what is h and k? You also have a point on the circle. What then, in addition to knowing the center, can you do with that? (Hint: Plug (0, 0) into the circle equation, or note that the line between (0, 0) and (-4, 3) is a radius.)

-Dan

3. Originally Posted by homeylova223
Write the equation of the circle that satisfies each set of conditions.

1.The endpoints of the diameters are (5,6) and (2,3)?

2. The circle passes through origin and has its center at (-4,3).

Now I know the standard form of the equation of a circle is

(x-h)^2+(y-k)^2=r^2

But how would I solve these two problems I mean for number 2 I know x and y but no the radius.
For the first note use the distance formula to find the length of the diamater.

$d=\sqrt{(6-3)^2+(5-2)^2}=\sqrt{3^2+3^2}=3\sqrt{2}$

So the radius is half of the diameter we get that

$r=\frac{1}{2}(3\sqrt{2})=\frac{3\sqrt{2}}{2} \implies r^2=\frac{9}{2}$

and the center is at the midpoint so we get

$x=\frac{5+2}{2}=\frac{7}{2} \quad y=\frac{6+3}{2}=\frac{9}{2}$

For the 2nd one use the distance formula to find the radius then just plug into the formula