# Math Help - Circles and Chords

1. ## Circles and Chords

Below, I have posted a question and a solution I was provided. The problem is that I don't understand how the part bordered by AAA was arrived at. Specifically $\sqrt{m^2+1}$, & why it's simply multiplied against $|3m+4|$? After multiplying that expression, I end up with $(3m^2+24m+16)(m^2+1)=20$, & not $(11m-2)(m-2)=0$. Can anyone help clarify this for me, or is there a better approach?

Many thanks.

Q. $x^2+y^2+2x-4y-20=0$ is the equation of a circle. 2 lines, L & M, intersect at (2, -2). The distance from the centre of the circle to each line is $2\sqrt{5}$. Find the equation of L & M.

Attempt: Line equation for (2,-2): $y+2=m(x-2)$
From circle equation: centre c = (-1, 2)
Sub c into the line equation: $2+2=m(-1-2)\rightarrow 3m+4=0$

AAA
The distance from the point (-1, 2) to this line is $|3m+4|\sqrt{m^2+1}=2\sqrt{5}$
$(11m-2)(m-2)=0$
AAA

Ans. (From text book): $2x-y-6=0$ & $2x-11y-26=0$

2. ## Re: Circles and Chords

Originally Posted by GrigOrig99
Below, I have posted a question and a solution I was provided. The problem is that I don't understand how the part bordered by AAA was arrived at. Specifically $\sqrt{m^2+1}$, & why it's simply multiplied against $|3m+4|$? After multiplying that expression, I end up with $(3m^2+24m+16)(m^2+1)=20$, & not $(11m-2)(m-2)=0$. Can anyone help clarify this for me, or is there a better approach?
Attempt: Line equation for (2,-2): $y+2=m(x-2)$
From circle equation: centre c = (-1, 2)
Sub c into the line equation: $2+2=m(-1-2)\rightarrow 3m+4=0$

AAA
The distance from the point (-1, 2) to this line is $|3m+4|\sqrt{m^2+1}=2\sqrt{5}$
$(11m-2)(m-2)=0$AAA

Ans. (From text book): $2x-y-6=0$ & $2x-11y-26=0$

The equations of the lines are $mx-y-2-2m=0$.

So the distance is $\frac{|-m-2-2-2m|}{\sqrt{m^2+1}}=2\sqrt{5}$ or $\frac{(3m+4)^2}{m^2+1}}=20$

3. ## Re: Circles and Chords

Thank you very much.

4. ## Re: Circles and Chords

I don't think that this is necessarily a better approach, but it is different.

The circle has a centre at $(-1,2)$ and has a radius of 5. If we draw a second smaller circle with the same centre and with radius $2\sqrt{5},$ then what we are looking for are the two tangents from the point $(2,-2),$ (which incidently lies on the larger circle), to the smaller one.

The smaller circle will have equation $(x+1)^{2}+(y-2)^{2}=(2\sqrt{5})^{2},$ or,
$x^{2}+y^{2}+2x-4y-15=0,$
and a tangent to this circle at the point $(x_{1},y_{1})$ will have equation
$xx_{1}+yy_{1}+(x+x_{1})-2(y+y_{1})-15=0.$

This, (these), has (have) to pass through the point $(2,-2),$ so $3x_{1}-4y_{1}-9=0.$

The point $(x_{1},y_{1})$ lies on the smaller circle, so substituting $x_{1}=(4y_{1}+9)/3,$ allows us to find the $y$ co-ordinates of the two points of contact. These turn out to be $0 \text{ and } -12/5.$
Corresponding $x$ co-ordinates are $3 \text{ and } -1/5.$

The equations of the two tangents can then be calculated (and agree with the given answers).