How do you write an equation for a circle with center (2,4) and passes through point (5, 9)?

1 Answer
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Apr 16, 2016

(x-2)^2 + (y-4)^2 = 34(x−2)2+(y−4)2=34

Explanation:

The radius can be found using Pythagoras Theorem

r = sqrt{(5-2)^2 + (9-4)^2} = sqrt34r=√(5−2)2+(9−4)2=√34

For a circle with radius rr and centered at (a,b)(a,b), its cartesian equation is

(x-a)^2 + (y-b)^2 = r^2(x−a)2+(y−b)2=r2

Therefore. for a circle with radius sqrt34√34 and centered at (2,4)(2,4). its cartesian equation is

(x-2)^2 + (y-4)^2 = 34(x−2)2+(y−4)2=34

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