Asked by Paige Pressler on Jul 15, 2024

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Write the standard form of the equation of the circle with the center at (0,0) that passes through the point (6,2) .

A) x2+y2=8x ^ { 2 } + y ^ { 2 } = 8x2+y2=8
B) x2+y2=40x ^ { 2 } + y ^ { 2 } = 40x2+y2=40
C) 4y2=36x24 y ^ { 2 } = 36 x ^ { 2 }4y2=36x2
D) 36y2=4x236 y ^ { 2 } = 4 x ^ { 2 }36y2=4x2
E) (x−6) 2+(y−2) 2=40( x - 6 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 40(x6) 2+(y2) 2=40

Center

In geometry, the central point of a circle or sphere, equidistant from all points on the perimeter or surface; in other contexts, it refers to a focal point or main area of activity.

Point

A location in space, often defined by its coordinates in a given dimension.

  • Enhance skill in devising and apprehending the normative models of equations related to circles.
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Stephanie ArroyoJul 15, 2024
Final Answer :
B
Explanation :
The standard form of the equation of a circle with center (h,k) and radius r is
(x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2(xh)2+(yk)2=r2
Since the center of the circle is (0,0), we have:
(x−0)2+(y−0)2=r2(x-0)^2+(y-0)^2=r^2(x0)2+(y0)2=r2
Simplifying, we have:
x2+y2=r2x^2+y^2=r^2x2+y2=r2
We need to find r. Since the circle passes through (6,2), this point is on the circle and satisfies the equation above. Substituting the values of x and y, we get:
62+22=r26^2+2^2=r^262+22=r2
Simplifying, we get:
40=r240=r^240=r2
Therefore, the equation of the circle is:
x2+y2=40x^2+y^2=40x2+y2=40
so the answer is B.