What is the equation of a circle centered at (2, 5) and passes through the point (4, 10)?

Answer 1

Hey Larson,

The left side of the equation of the circle is easy to plot from the Center (2,5) using the standard form for circles: (x-h)^2 + (y-k)^2 = r^2 the (x,y) values at the center point are your h and k values, so

we have (x-2)^2 + (y-5)^=r^2 now we need the radius of the circle to finish the equation.

The circle goes through the point (4,10) with the center at (2,5). That means the radius covers the distance of 5 up and 2 over. If we use pythagorean theorem we can find the radius from the hypotenuse based of a right triangle with base and height legs of 2 and 5. 2^2 + 5^2 = r^2. 4 + 25 = 29 = r^2, so the radius equals the square root of 29. Plug that value in place of r and we get

(X-2)^2 + (Y-5)^2 = 29

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Answer 2

Larson,

To find the equation of this circle, we will use the general equation of a circle (which is just distance formula):

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

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

To write this equation, we are plugging numbers in for h, k, and r, to leave y and x varying. However, we do not know what the radius is, but we do know where the center is and another point on the circle. So, we can use the equation to find r^2 (the right side of the equation).

If we plug in (2, 5) for h and k since it's the center and (4, 10) for x and y since it is another point on the circle:

(4 - 2)^2 + (10 - 5)^2 = r^2

Simplify:

(2)^2 + (5)^2 = r^2

---> 4 + 25 = r^2

---> 29 = r^2

Now, we can plug in for h, k, and r^2.

So, the equation of this circle is (x - 2)^2 +(y - 5)^2 = 29

Hope this helps