Find an equation of the circle and sketch it: i Radius 5, tangent to y-axis; contain (10, 7)

Answer 1

To be tangent to y-axis and a radius of 5, the center must be either (5, y) or (-5, y).

(x - 5)² + (y - k)² = 25

The circle contains (10, 7)

(10 - 5)² + (7 - k)² =25

25 = 49 - 14k + k² = 25

0 = k² - 14k + 49

0 = (k - 7)²

7 = k

Answer 2

The generic equation for a circle is (x - h)² + (y - k)² = r² where (h, k) is the center of the circle. But if the circle is tangent to the y axis with a radius of 5, then the center must be somewhere along the x = 5 line. So, we can now write the equation as (x - 5)² + (y - k)² = 25. We just need to know what "k" is.

But we know that the circle goes through the point (10, 7) so we can plug that in as an (x, y). So: (10- 5)² + (7 - k)² = 25

Multiplying this out gives:

25 + (49 - 14k + k²) = 25

or k² - 14k + 49 = 0

or (k - 7)(k - 7) = 0

or k = 7

So we can write the equation of the circle as (x - 5)² + (y - 7)² = 25