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Princess' question at Yahoo! Answers: finding lines through the origin and tangent to given circle

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MarkFL

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Feb 24, 2012
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Here is the question:

How to find equations of tangent to a circle?


Find the equations of tangents to

x^2+y^2-6x-2y+9=0

through the origin. Also find their respective points of contact.

thanks
I have posted a link there to this thread so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Princess,

We are given the equation of the circle:

\(\displaystyle x^2+y^2-6x-2y+9=0\)

The family of lines through the origin is given by:

\(\displaystyle y=mx\) where \(\displaystyle m\in\mathbb{R}\)

Substituting for $y$, we obtain the following quadratic in $x$:

\(\displaystyle x^2+(mx)^2-6x-2(mx)+9=0\)

Arranging in standard form, we have:

\(\displaystyle \left(m^2+1 \right)x^2-2(3+m)x+9=0\)

Since the line $y=mx$ is tangent to the circle, the discriminant must be zero:

\(\displaystyle \left(-2(3+m) \right)^2-4\left(m^2+1 \right)(9)=0\)

\(\displaystyle 9+6m+m^2-9m^2-9=0\)

\(\displaystyle 8m^2-6m=0\)

\(\displaystyle m(4m-3)=0\)

\(\displaystyle m=0,\,\frac{3}{4}\)

Thus, the tangent lines are:

\(\displaystyle y=0\)

\(\displaystyle y=\frac{3}{4}x\)

Here is a plot of the circle and the two tangent lines:

princess.jpg

To find the contact points, we will substitute for $y$ into the circle:

i) \(\displaystyle y=0\)

\(\displaystyle x^2+0^2-6x-2(0)+9=0\)

\(\displaystyle (x-3)^2=0\)

\(\displaystyle x=3\)

This contact point is \(\displaystyle (3,0)\).

ii) \(\displaystyle y=\frac{3}{4}x\)

\(\displaystyle x^2+\left(\frac{3}{4}x \right)^2-6x-2\left(\frac{3}{4}x \right)+9=0\)

\(\displaystyle \frac{25}{16}x^2-\frac{15}{2}x+9=0\)

\(\displaystyle 25x^2-120x+144=0\)

\(\displaystyle (5x-12)^2=0\)

\(\displaystyle x=\frac{12}{5}\implies y=\frac{9}{5}\)

This contact point is \(\displaystyle \left(\frac{12}{5},\frac{9}{5} \right)\).