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

1 Answer

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Answer:

(x+10)² + (y+5)² = 125

Explanation:

Pre-Solving

We are given that a circle has a center (-10,-5), and passes through the point (-5,5).

We want to write the equation of this circle in the standard equation. The standard equation is (x-h)² + (y-k)² = r² where (h,k) is the center and r is the radius.

Solving

As we are already given the center point, we can substitute its values into the equation.

Reminder: the equation uses negative values, and we have negative numbers.

(x--10)² + (y--5)² = r²

This can be simplified to:

(x+10)² + (y+5)² = r²

Now, we need to find r².

As the point passes through (-5,5), we can use its values to solve for r².

Substitute -5 as x and 5 as y.

(-5+10)² + (5+5)² = r²

(5)² + (10)² = r²

25 + 100 = r²

125=r²

The radius is 125

Substitute 125 as r².

(x+10)² + (y+5)² = 125

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User Mickl
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