Question

quiz equations of tangents to a circle
find the equation of the tangent to the circle $x^2 + y^2 - 12x + 6y - 35 = 0$ at the point $(-2, 1)$.
give your answer in the form $y = mx + c$ where $m$ and $c$ are constants.
show your working.

Explanation:

Step1: Find the center of the circle

First, we rewrite the circle equation \(x^{2}+y^{2}-12x + 6y-35 = 0\) in standard form \((x - a)^{2}+(y - b)^{2}=r^{2}\), where \((a,b)\) is the center and \(r\) is the radius.
Completing the square for \(x\) and \(y\) terms:
For \(x\): \(x^{2}-12x=(x - 6)^{2}-36\)
For \(y\): \(y^{2}+6y=(y + 3)^{2}-9\)
Substituting back into the equation:
\((x - 6)^{2}-36+(y + 3)^{2}-9-35 = 0\)
Simplify: \((x - 6)^{2}+(y + 3)^{2}=80\)
So the center of the circle \(C\) is \((6,-3)\).

Step2: Find the slope of the radius

The radius is the line joining the center \(C(6,-3)\) and the point of tangency \(P(-2,1)\).
The slope of the radius \(m_{CP}\) is given by the formula \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Substituting \(x_{1}=6,y_{1}=-3,x_{2}=-2,y_{2}=1\):
\(m_{CP}=\frac{1-(-3)}{-2 - 6}=\frac{4}{-8}=-\frac{1}{2}\)

Step3: Find the slope of the tangent

The tangent to a circle at a point is perpendicular to the radius at that point. If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the tangent be \(m_{t}\).
We have \(m_{CP}\times m_{t}=-1\)
Substituting \(m_{CP}=-\frac{1}{2}\):
\(-\frac{1}{2}\times m_{t}=-1\)
Solving for \(m_{t}\), we get \(m_{t}=2\)

Step4: Find the equation of the tangent

We use the point - slope form of a line \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(-2,1)\) and \(m = m_{t}=2\)
Substituting the values:
\(y - 1=2(x+2)\)
Expand the right - hand side: \(y - 1=2x + 4\)
Add 1 to both sides: \(y=2x+5\)

Answer:

\(y = 2x+5\)