# Question #9a3da

Sep 23, 2015

$\left(a\right) . {66.84}^{\circ}$

$\left(b\right) . \left(\frac{13}{4} , \frac{1}{2}\right)$

#### Explanation:

${x}^{2} + {y}^{2} - y - 3 = 0$

As you can see this describes a circle:

graph{x^2+y^2-y-3=0 [-10, 10, -5, 5]}

The 2 tangents where the line $x = 1$ cuts the circle fit the general equation of a straight line:

$y = m x + c$

We can find $m$ and $c$ to get the equations of the tangents and then find the angle of intersection.

To get the gradients, we can differentiate both sides implicitly since both $x$ and $y$ appear together:

$D \left({x}^{2} + {y}^{2} - y - 3\right) = D \left(0\right)$

$2 x + 2 y y ' - y ' = 0$

$y ' \left(2 y - 1\right) = - 2 x$

$y ' = \frac{- 2 x}{2 y - 1}$

If $x = 1$ then that line will cut the circle giving 2 values for $y$:

$1 + {y}^{2} - y - 3 = 0$

${y}^{2} - y - 2 = 0$

Factorising:

$\left(y + 1\right) \left(y - 2\right) = 0$

If $\left(y + 1\right) = 0$

$y = - 1$

If $\left(y - 2\right) = 0$

$y = 2$

So these are the 2 values of $y$ where the $x = 1$ line cuts the circle.

Now to get the equation of the 1st tangent where $x = 1$ and $y = - 1$:

$m = y ' = \frac{- 2 x}{2 y - 1} = \frac{- 2 \times 1}{2 \times \left(- 1\right) - 1} = \frac{- 2}{- 3} = \frac{2}{3}$

$m = \frac{2}{3}$

To get $c$:

$y = m x + c$:
$- 1 = \frac{2}{3} \times 1 + c$

$c = - 1 - \frac{2}{3} = - \frac{3}{3} - \frac{2}{3} = - \frac{5}{3}$

The equation for the tangent $\Rightarrow$

$y = \frac{2}{3} x - \frac{5}{3} \text{ } \textcolor{red}{\left(1\right)}$

Now to get the 2nd tangent:

$x = 1 , y = 2$

$m = y ' = \frac{- 2 x}{2 y - 1} = \frac{- 2 \times 1}{\left(2 \times 2\right) - 1}$

$m = - \frac{2}{3}$

To get $c$:

$y = m x + c$

$2 = - \frac{2}{3} \times 1 + c$

$c = 2 + \frac{2}{3} = \frac{6}{3} + \frac{2}{3} = \frac{8}{3}$

The equation becomes:

$y = - \frac{2}{3} x + \frac{8}{3} \text{ } \textcolor{red}{\left(2\right)}$

At the intersection of the 2 tangents we can put $\textcolor{red}{\left(1\right)}$ equal to $\textcolor{red}{\left(2\right)} \Rightarrow$

$\frac{2}{3} x - \frac{5}{3} = - \frac{2}{3} x + \frac{8}{3}$

$\frac{2 x}{3} + \frac{2 x}{3} = \frac{8}{3} + \frac{5}{3}$

$\frac{4 x}{3} = \frac{13}{3}$

$x = \frac{13}{4}$

Now to get the $y$ co-ordinate:

$y = \frac{2 x}{3} - \frac{5}{3}$

$y = \frac{2}{3} \times \frac{13}{4} - \frac{5}{3}$

$y = \frac{26}{12} - \frac{5}{3} = \frac{26}{12} - \frac{20}{12} = \frac{6}{12} = \frac{1}{2}$

$y = \frac{1}{2}$

So the (x,y) co-ordinates of the intersection are $\left(\frac{13}{4} , \frac{1}{2}\right)$

To get the angle of intersection between the 2 tangents there is an expression you can use but I think its best to look at the geometry to see what's going on:

Here you can see that the angle of intersection is ${66.84}^{\circ}$

Oct 2, 2015

Michael has given a fine answer using a bit of calculus. Here is a partial answer without calculus.

#### Explanation:

The circle ${x}^{2} + {y}^{2} - y - 3 = 0$ can be put into standard form:

${x}^{2} + {\left(y - \frac{1}{2}\right)}^{2} = \frac{13}{4}$

The center of the circle is $\left(0 , \frac{1}{2}\right)$.

The points on the circle with $x = 1$ are $\left(1 , 2\right)$ and $\left(1 , - 1\right)$

To find the slopes of the tangent lines at these points, use the fact that a tangent to a circle at a point is perpendicular to the radius at that point.

At $\left(1 , - 1\right)$, the slope of the radius is $\frac{- 1 - \left(\frac{1}{2}\right)}{1 - 0} = - \frac{3}{2}$

So the slope of the tangent at $\left(1 , - 1\right)$ is $\frac{2}{3}$
See Michael's answer to get the equation of the tangent line at $\left(1 , - 1\right)$ is
$y = \frac{2}{3} x - \frac{5}{3} \text{ } \textcolor{red}{\left(1\right)}$

At $\left(1 , 2\right)$, the slope of the radius is $\frac{2 - \left(\frac{1}{2}\right)}{1 - 0} = \frac{3}{2}$

So the slope of the tangent at $\left(1 , 2\right)$ is $- \frac{2}{3}$
See Michael's answer to get the equation of the tangent line at $\left(1 , 2\right)$ is
$y = - \frac{2}{3} x + \frac{8}{3} \text{ } \textcolor{red}{\left(2\right)}$

For the remainder of the solution, see Michael's answer.
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