Circles on the Coordinate Plane Examples

Give an equation for the circle with center (-14, 54) and radius 64 units.

The implicit formula for a circle with center (h, k) and radius r is (x – h)² + (y – k)² = r². We're given the center and radius, so all we need to do is plug the information into the right places. 

(x – h)² + (y – k)² = r²
(x – (-14))² + (y – 54)² = 64² 
(x + 14)² + (y – 54)² = 64² 

It's okay to keep the 64² in that form.

Give an equation for the circle with center (107, -3.78) and radius 4 units.

The implicit formula for a circle with center (h, k) and radius r is (x – h)² + (y – k)² = r². Just plug 'em and chug 'em:

(x – h)² + (y – k)² = r² 
(x – 107)² + (y – (-3.78))² = 4² 
(x – 107)² + (y + 3.78)² = 16

Is the point (-10, 30) on the circle with equation (x + 15)² + (y – 35)² = 102? If not, is the point in the interior or exterior of the circle?

Remember that the equation is set up so that every point (x, y) that makes the equation true is on the circle (and every point on the circle makes the equation true). To find out whether (-10, 30) is on the circle with equation (x + 15)² + (y – 35)² = 102, we just plug in -10 for x and 30 for y and see whether the equation is true.

((-10) + 15)² + ((30) – 35)² ≟ 102
(5)² + (-5)² ≟ 102
25 + 25 ≟ 102
50 ≠ 102, so (-10, 30) is not on the circle.

Also remember that while the right-hand side of the formula is the square of the radius, the left-hand side is the square of the distance between (x, y) and the center of the circle. In this case, we have 50 < 102. Since , we know that (-10, 30) is in the interior of the circle.

Is the point (20, -12) on the circle with equation (x – 17)² + (y + 8)² = 25? If not, is the point in the interior or exterior of the circle?

Remember that the equation is set up so that every point (x, y) that makes the equation true is on the circle (and every point on the circle makes the equation true). To find out whether (20, -12) is on the circle with equation (x – 17)² + (y + 8)² = 25, we just plug in 20 for x and -12 for y and see whether the equation is true.

((20) – 17)² + ((-12) + 8)² ≟ 25
3² + (-4)² ≟ 25
9 + 16 ≟ 25
9 + 16 = 25, so the point (20, -12) is on the circle.

Given the circle with equation (x + 22)² + (y – 56)² = 1681, give an equation for the line tangent to the circle at the point (-13, 96).

First, find the slope of the radius of the circle at the point of tangency. Here, the point of tangency is (-13, 96) and the center of the circle is (-22, 56). So the slope of the radius is given by

The slope of the tangent line (which is always perpendicular to the radius) is the negative reciprocal, or . Now we have enough information to write an equation for the tangent line in point-slope form: .

Given the circle with equation (x – 14)² + (y + 3)² = 169, give an equation for the line tangent to the circle at the point (19, 9).

Okay, so first off, our point of tangency is (19, 9) and the center of the circle is (14, 3). Plug those guys into the slop formula to get the slope of the radius:

The slope of the tangent line (which is always perpendicular to the radius) is the negative reciprocal, or . Now plop all that info into the point-slope formula: