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Write the standard equation of a circle that passes through (−5, 5) with center (−10, −5).

A: (x – 10)2^+ (y – 5)2^= 75

B: (x + 10)2^+ (y + 5)2^= 125

C: (x + 10)2^+ (y + 5)2^= 75

D: (x – 10)2^+ (y – 5)2^= 125

Respuesta :

grujan50

Answer:

Correct answer:  B: (x + 10)² + (y + 5)² = 125

Step-by-step explanation:

Given:

Point A(-5, 5)   point that belong to the circle

Point C(-10, -5) = (a, b)   coordinates of the center of the circle

The standard form of the circle equation is:

(x - a)² + (y - b)² = r²

we will replace the given coordinates in the circle equation

(-5 -(-10))² + (5 -(- 5))² =  r²

(-5 + 10)² + (5 + 5)² =  r²

5² + 10² = 125 = r²

r² = 125

the required equation of the circle is:

(x + 10)² + (y + 5)² = 125

God is with you!!!

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