how do i find the equations of L1 and L2

Question

L1 and L2 are perpendicular. the equation of the circle is given as $x^2+6x+y^2-2y=7$. line L1 cuts the circle at $P$. L2 cuts the circle at $Q$. I need to find the equations of the lines L1 and L2. I tried to do this by getting the equation parallel to L2 that passes through $P$ and the centre $(-3,1)$. so $AP$ gradient * L2 gradient$= -1$. and for the second equation to solve the simultaneous I took the length $AP$ so $17= (Px+3)^3 + (Py -1)^2$.. the answer I get when I do this is not a rational number while the answer is a rational number. How do I handle this question?

Answer 1

We need to determine the equations for lines L1 and L2. You have attempted to use the product of gradients to determine the equation for L2, and solved it using a length AP of 17.

Firstly, let's take a look at how to find the equation for line L1. According to the problem description, we know that line L1 passes through point P and the tangent point of the circle. Therefore, we need to find the coordinates of point P and the slope of L1.

For the equation $ x ^ 2+6x+y ^ 2-2y=7$ of a circle, converting it to its standard form yields:

$(x+3) ^ 2-9+(y-1) ^ 2-1=7 $

$(x+3) ^ 2+(y -1) ^ 2=17 $

Since point P is on L1 and cuts a circle, it must also satisfy the equation for the circle. Therefore, we can substitute the coordinates of point P into the equation of the circle, and then solve the system of equations to find the coordinates of point P.

Next, we can use point P and the center point of the circle (-2, 3) to calculate the slope of line L1. Since L1 is perpendicular to L2, we can calculate the slope of L2 by taking the negative reciprocal of the slope.

Now that we have found the slopes of lines L1 and L2, we can use point oblique or general expressions to represent the equations of these two lines.

As for the condition where the length AP is 17, in this case, you may need to use the distance formula from the point to the center of the circle instead. The formula is: $(Px - (-2)) ^ 2+(Py -3) ^ 2=17$ In this way, you can obtain equations containing L1 and L2.

Answer 2

If you notice here that because the 2 tangents are perpendicular, the quadrilateral made by them and the 2 respective radii at those point of contacts in fact form a square.

We know that the diagonal acts as an angle bisector. Here we will use that. The centre of circle and point of intersection of tangents will act as diagonal.

Calculating slope between these 2 points, we get m or $\tan \theta = \frac {5}{3} $

Let us assume the slope of side of square to be M or $ \tan \phi $

$$\left| \phi - \theta \right| = \frac {π}{4} $$

Applying difference of tangent identity i.e. $$ \tan (\theta - \phi) = \left| \frac {\tan \theta - \tan \phi}{1 + \tan\theta \cdot \tan\phi} \right| $$

We can do $\phi - \theta$ and obtain another value but $\tan$ is an odd function so we can take the - out. Hence I have used absolute value there.

Evaluating the values we get $\tan \phi$ = $0.25,-4$

We have our point $(0,6)$ and know the slopes so we can get L1 and L2.

Edit - 1 Upon clicking post, I got a much simpler way to solve this. I wont write another answer so I will just summarize it here.

Assume the tangent lines to be $y = mx + c$

They are passing from $(0,6)$. Satisfying them into the line, we get c. Then we know perpendicular distance from tangent to centre is radius of circle. Using the formula of perpendicular distance, we can get a quadratic whose roots are the required slopes.

Hope this helps.

Edit - 2 Elaborating on the other method I mentioned in the previous edit as requested by OP.

As $ (0,6) $ lies on the y-axis. We can directly say that %c = 6%. So our line becomes $ y = mx + 6 $. We just require m now.

The formula of perpendicular distance of a line $ ax+by+c = 0$ from point $P(h,k)$ is given by

$$ d(P,r)= \frac {|ah+bk+c|}{\sqrt {a^2 + b^2} } $$

We know that perpendicular distance of a tangent line from the centre of a circle is equal to the radius of the circle. So the equation will be

$$ \frac {|-3m-1+6|}{\sqrt {1^2 + m^2} } = \sqrt {17} $$

Take the denominator to the RHS and square it to obtain the quadratic equation.

$$ 4m^2 + 15m -4 = 0 $$

Which nicely factors if you write $15m$ as $16m - m$ and yields

$$ (4m-1)(m+4) = 0 $$ $$ m = \frac {1}{4}, 4 $$

Applying point slope form will complete the answer.