Find the equation of tangent to the circle x2 + y2 + 2x + 4y − 12 = 0 at (3, −1)

Question

ScanSolve AI Tutor · Accepted Answer

To find the equation of the tangent to the circle $x^2 + y^2 + 2x + 4y - 12 = 0$ at the point $(3, -1)$ , we will first find the center of the circle and then use the fact that the tangent is perpendicular to the radius at the point of tangency.

Find the center of the circle.
The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$ , where the center is $(-g, -f)$ .
Comparing the given equation $x^2 + y^2 + 2x + 4y - 12 = 0$ with the general form:
$2g = 2 \implies g = 1$
$2f = 4 \implies f = 2$
The center of the circle is $C(-g, -f) = C(-1, -2)$ .
Find the slope of the radius.
The radius connects the center $C(-1, -2)$ to the point of tangency $P(3, -1)$ .
The slope of the radius $m_r$ is given by:
$m_r = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - (-2)}{3 - (-1)} = \frac{-1 + 2}{3 + 1} = \frac{1}{4}$
Find the slope of the tangent line.
The tangent line is perpendicular to the radius at the point of tangency. If $m_t$ is the slope of the tangent, then $m_t \cdot m_r = -1$ .
$m_t \cdot \frac{1}{4} = -1$ $m_t = -4$
Find the equation of the tangent line.
Using the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$ , with the point $P(3, -1)$ and slope $m_t = -4$ :
$y - (-1) = -4(x - 3)$ $y + 1 = -4x + 12$ $y = -4x + 12 - 1$ $y = -4x + 11$
Rearranging into the standard form $Ax + By + C = 0$ :
$4x + y - 11 = 0$

Find the equation of tangent to the circle x2 + y2 + 2x + 4y − 12 = 0 at (3, -1)

Answer

Step-by-step solution

Find the center of the circle.

Find the slope of the radius.

Find the slope of the tangent line.

Find the equation of the tangent line.

Try one yourself

Still stuck on this one?

More Mathematics Questions