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The equation for the circle is:

x2+y2+14x+10y−7=0 .



What is the center of the circle?

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User Shfire
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1 Answer

4 votes

Answer:

(-7, -5)

Explanation:

x² + y² + 14x + 10y − 7 = 0

Strategy:

Convert the equation to the centre-radius form:

(x - h)² + (y - k)² = r²

The centre of the circle is at (h, k) and the radius is r

Solution:

Group x and y terms together; move the number to the right-hand side.

x² + 14x + y² + 10y = 7

Complete the square for x

(Take half the coefficient of x, square it, and add to each side of the equation)

(14/2)² = 7² = 49

(x² + 14x + 49) + (y² + 10y) = 56

Complete the square for y

(Take half the coefficient of y, square it, and add to each side of the equation)

(10/2)² = 5² =25

(x² + 14x + 49) + (y² + 10y + 25 ) = 81

Express the result as the sum of squares

(x + 7)² + (y + 5)² = 9²

h = -7; k = -5; r = 9

The centre of the circle is at (-7, -5).

The graph of the circle below has its centre at (-7, -5) and radius 9.

The equation for the circle is: x2+y2+14x+10y−7=0 . What is the center of the circle-example-1
User Aleksei Poliakov
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5.0k points