# Circles and Chords

• May 20th 2013, 12:58 PM
GrigOrig99
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 $\displaystyle \sqrt{m^2+1}$, & why it's simply multiplied against $\displaystyle |3m+4|$? After multiplying that expression, I end up with $\displaystyle (3m^2+24m+16)(m^2+1)=20$, & not $\displaystyle (11m-2)(m-2)=0$. Can anyone help clarify this for me, or is there a better approach?

Many thanks.

Q. $\displaystyle x^2+y^2+2x-4y-20=0$ 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 $\displaystyle 2\sqrt{5}$. Find the equation of L & M.

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

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

Ans. (From text book): $\displaystyle 2x-y-6=0$ & $\displaystyle 2x-11y-26=0$
• May 20th 2013, 01:19 PM
Plato
Re: Circles and Chords
Quote:

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 $\displaystyle \sqrt{m^2+1}$, & why it's simply multiplied against $\displaystyle |3m+4|$? After multiplying that expression, I end up with $\displaystyle (3m^2+24m+16)(m^2+1)=20$, & not $\displaystyle (11m-2)(m-2)=0$. Can anyone help clarify this for me, or is there a better approach?
Attempt: Line equation for (2,-2): $\displaystyle y+2=m(x-2)$
From circle equation: centre c = (-1, 2)
Sub c into the line equation: $\displaystyle 2+2=m(-1-2)\rightarrow 3m+4=0$

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

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

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

So the distance is $\displaystyle \frac{|-m-2-2-2m|}{\sqrt{m^2+1}}=2\sqrt{5}$ or $\displaystyle \frac{(3m+4)^2}{m^2+1}}=20$
• May 20th 2013, 01:34 PM
GrigOrig99
Re: Circles and Chords
Thank you very much.
• May 22nd 2013, 03:57 AM
BobP
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 $\displaystyle (-1,2)$ and has a radius of 5. If we draw a second smaller circle with the same centre and with radius $\displaystyle 2\sqrt{5},$ then what we are looking for are the two tangents from the point $\displaystyle (2,-2),$ (which incidently lies on the larger circle), to the smaller one.

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

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

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

The equations of the two tangents can then be calculated (and agree with the given answers).
• May 22nd 2013, 07:21 PM
ibdutt
Re: Circles and Chords