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Write the standrd form of the equation of a circle that contains the point (6,15) and has center (12,12).

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User Myxtic
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1 Answer

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Final answer:

The standard form of the equation of a circle that contains the point (6,15) and has center (12,12) is (x - 12)² + (y - 12)² = 45.

Step-by-step explanation:

The standard form of the equation of a circle with center (h, k) and radius r is given by:

(x - h)² + (y - k)² = r²

To find the standard form of the equation of a circle that contains the point (6,15) and has center (12,12), we first need to calculate the radius of the circle. The radius can be found by using the distance formula between the center and the point on the circle:

r = √((x2 - x1)² + (y2 - y1)²)

Substituting the coordinates:

r = √((6 - 12)² + (15 - 12)²) = √((-6)² + (3)²) = √(36 + 9) = √45

Now we use the center (h, k) = (12, 12) and r = √45 to write the standard form of the equation:

(x - 12)² + (y - 12)² = (√45)²

(x - 12)² + (y - 12)² = 45

This is the standard form of the equation for the circle requested.

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User NMKP
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