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Find the equation of the circle passing through the point of intersection of x²+y²−2x−4y−4=0.

a. x²+y²−2x−4y−4=0
b. x²+y²+2x+4y+4=0
c. x²+y²+2x−4y−4=0
d. x²+y²−2x+4y+4=0

User Scarlz
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1 Answer

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Final answer:

The equation of the circle passing through the point of intersection of x²+y²−2x−4y−4=0 can be found by completing the square and rearranging the terms. The equation becomes (x - 1)² + (y - 2)² = 9. The correct option is d. x²+y²−2x+4y+4=0.

Step-by-step explanation:

The equation of the circle passing through the point of intersection of x²+y²−2x−4y−4=0 can be found by completing the square and rearranging the terms.

The equation can be rewritten as (x² - 2x) + (y² - 4y) = 4.

Completing the square for x and y by adding the square of half the coefficient of x and y, we get (x² - 2x + 1) + (y² - 4y + 4) = 4 + 1 + 4.

Combining like terms, the equation becomes (x - 1)² + (y - 2)² = 9.

Therefore, the equation of the circle passing through the point of intersection is (x - 1)² + (y - 2)² = 9, which is the same as option d: x²+y²−2x+4y+4=0.

The question is to find the equation of the circle passing through the point of intersection of x²+y²−2x−4y−4=0. However, the given point of intersection cannot determine a unique circle without some additional information, such as another point of intersection or the circle's radius.

User Larry K
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