Question

How do you write an equation for a circle tangent to the line 5x + y = 3 at the point (2,-7) and the center is on the line x – 2y= 19?

Answer 1

Please see the explanation. The equation is:
`(x - 7)^2 + (y - -6)^2 = (sqrt(26))^2`

Explanation:

Write the equation of the tangent line in slope-intercept form so that we may observe its slope.

`y = -5x + 3`

The slope `m = -5`

Because radius of the circle at point (2, -7) will be perpendicular to the tangent line, the slope of the radius is the negative reciprocal of the slope of the tangent line, `1/5`, Write the equation of the line through the point with this slope:

`-7 = 1/5(2) + b`

`b = -7 -2/5`

`b = -37/5`

`y = 1/5x - 37/5` [1]

The center `(h, k)` will be at the intersection of equation [1] and the line:

`x - 2y = 19` [2]

Substitute equation [1] into equation [2]:

`x - 2(1/5x - 37/5) = 19`

`x - 2/5x + 74/5 = 19`

`3/5x = 21/5`

`x = 7`

Substituting for x in equation [2]:

`7 - 2y = 19`

`- 2y = 12`

`y = -6`

Substitute the center `(7, -6)` and the point `(2, -7)` into the equation of a circle:

`(2 - 7)^2 + (-7 - -6)^2 = r^2`

#r = sqrt(26)