Question

How to find the equation for a tangent to a circle with only one point?

How do you find the equation for a tangent to the circle `x^2+y^2-2x+10y-19=0` that passes through the point `(7,-2)`?

Answer 1

Please see the explanation.

Explanation:

Given `x^2+y^2-2x+10y-19=0` and the point `(7,-2)`

Verify that the point is on the circle:

`7^2+(-2)^2-2(7)+10(-2)-19=0`

`0=0` verified

Differentiate each term of the equation:

`(d(x^2))/dx+(d(y^2))/dx-(d(2x))/dx+(d(10y))/dx-(d(19))/dx=0`

`2x+2ydy/dx-2+10dy/dx-0=0`

Move the terms that do not contain `dy/dx` to the right side:

`2ydy/dx+10dy/dx=2-2x`

Factor the left side:

`(2y+10)dy/dx=2-2x`

Divide both sides by `2y+10`:

`dy/dx=(2-2x)/(2y+10)`

The slope, m, of the tangent line at the point `(7,-2)` is the above derivative evaluated at the point:

`m = (2-2(7))/(2(-2)+10)`

`m = (-12)/6`

`m = -2`

Use the point slope form of the equation of line:

`y = m(x - x_1)+y_1`

`y = -2(x - 7) - 2 larr` point-slope form

`y = -2x + 14 - 2`

`y = -2x + 12 larr` slope-intercept form

`2x + y = 12 larr` standard form