How do I find an equation for a circle with a center of (0,2) that passes throught he point (4,0)?

Question

Had this on a test and it completely stumped me. Since then, I've lost the math book that would give me the answer. I was hoping there'd be a math genius that'd know off hand. ^-^

Answer 1

You know the center, so the equation is in the form of
(x - 0)^2 + (y - 2)^2 = r^2

where "r" is the radius. We just need to know what r is.

We're not given the radius, but we know (4,0) is a point on the circle. So plug in (4,0) and solve for r, then use r to write the final equation.

(4)^2 + (-2)^2 = r^2
16 + 4 = r^2
20 = r^2

Come to think of it, there's actually no need to take the square root and then square it back. We know that the equation is:
(x)^2 + (y - 2)^2 = 20

Answer 2

The general eqn for a circle is (x-a)^2 + (y-b)^2 = c^2
where, (a,b) is the center of the circle and c is the radius of the circle.
You have a point on the circumference and the center, so you can get its radius.
c = sqrt[(4-0)^2 + (0-2)^2] = sqrt(16 + 4) = sqrt(20)
Now, substitute (0,2) and 2sqrt5 into the general eqn and then simplify it;
(x-0)^2 + (y-2)^2 = 20
x^2 + y^ - 4y + 4 = 20
x^2 + y^ - 4y - 16 = 0

Answer 3

Geezah's right.

use (y - 2)^2 + (x - 0)^2 = r^2
Radius (r) is distance of a point on the circle, i.e. (4,0) and the centre, (0,2). Then :
r^2 = (0 - 2)^2 + (4 - 0)^2
r^2 = 20

Then, the equation become
(y - 2)^2 + x^2 = 20

See?

Answer 4

Circle centre (a,b) and radius r has equation :-
(x - a)² + (y - b)² = r²

(x - 0)² + (y - 2) = 2² + 4²
x² + (y - 2)² = 20
Which may be written as:-
x² + y² - 4y + 4 = 20
x² + y² - 4y - 16 = 0

Answer 5

circle passing through (4, 0) with (0, 2) as centre is
(x -- 0)^2 + (y -- 2)^2 = (4 -- 0)^2 + (0 -- 2)^2
or x^2 + y^2 -- 4y -- 16 = 0

Answer 6

draw the circle and use a ruler!!!!!!! only way