Question

Find an equation for a circle satisfying the given conditions. (a) Center (-1, 4), passes through (3, 7) (b) The points (7, 13) and (-3, -11) are at the ends of a diameter.

Answer 1

Explanation:

In order to find the equations we need the circle's general equation:

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

(h,k) is the center and 'r' is the radius.

A. Because the center is (-1,4) then h=-1 and k=4.

Now we can find the radius as:

`distance=\sqrt{(x2-x1)^(2)+(y2-y1)^(2)}`

`distance=\sqrt{(3-(-1))^(2)+(7-4)^(2)}`

`distance=5` so we have r=5

Then the equation is
`(x+1)^(2)+(y-4)^(2)=25`

B. Because we have two points defining a diameter we can find the radius as follows:

`diameter=\sqrt{(x2-x1)^(2)+(y2-y1)^(2)}`

`diameter=\sqrt{(7-(-3))^(2)+(13-(-11))^(2)}`

`diameter=26`

`radius=26/2=13`

Now let's find the center of the circle as follows:

`C=((x1+x2)/(2) , (y1+y2)/(2))`

`C=((7-3)/(2) , (13-11)/(2))`

`C=(2,1)`

Then the equation is
`(x-2)^(2)+(y-1)^(2)=169`