Question

IVE

Write the standard form of the equation of the circle described below.
Center (0.3) passes through the point (-4,4)
The standard form of the equation of the circle is
(Type an equation. Simplify your answer.)

Answer 1

`x^2 +(y-3)^2=17` is the standard form of equation of circle

Solution:

Given that a circle has center (0, 3) and passes through point (-4, 4)

We have to find the standard form of equation of circle

The standard form of equation of circle is given as:

`(x-h)^2+(y-k)^2=r^2` --------- eqn 1

Where the center being at the point (h, k) and the radius being "r"

Here center = (h, k) = (0, 3)

Plug in (h, k) = (0, 3) and (x, y) = (-4, 4) in above equation

`(-4-0)^2+(4-3)^2=r^2\\\\16+1 = r^2\\\\r^2 = 17`

Now plug in
`r^2=17` and (h, k) = (0, 3) in eqn 1

`(x-0)^2+(y-3)^2=17\\\\x^2 +(y-3)^2=17`

Thus the standard form of equation is found