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Find the intersection point of the circles with equations

$$(x-3)^2+(y+4)^2=100$$

and

$$(x-15)^2+(y-12)^2=100.$$

I tried subtracting both equations but I don't know what to do next, any help appreciated.

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    when you subtract them, you will get a linear equation on $x$ and $y$. substitute it into one of the original two to get a quadratic equation in one variable. – 2017-02-14

3 Answers 3

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From these two equations, you get \begin{equation*} \begin{split} (x-3)^2+(y+4)^2=(x-15)^2+(y-12)^2 \Rightarrow \\ x^2-6x+9+y^2+8y+16=x^2-30x+225+y^2-24y+144 \Rightarrow \\ 24x+32y=344 \Rightarrow \\3x+4y=43 \end{split} \end{equation*}

Now, $x=\frac{43-4y}{3}$. Plug it in one of the equations, and solve it.

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Subtracting one from the other is a good idea, since the $x^2$ and $y^2$ terms will cancel (once you expand the brackets), leaving you with the linear equation

$$ 3x+4y=43 \iff x = \frac{43-4y}{3}$$

Substituting this value for $x$ into one of the original equations leaves you with a quadratic equation in $y$, namely

$$ y^2 - 8y + 16=0$$

Solving this gives you $y$, and then you can use the expression for $x$ in terms of $y$ to calculate $x$.

Obviously I have ommitted lots of the working out here, so I would recommend going through them yourself.

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    Thank you very much, Its solved :) – 2017-02-14
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Note that the distance between the centres is equal to the sum of the radii, which are both equal to $10$, so the point of intersection (in fact the point where they touch) is the midpoint of the line joining the centres, i.e. $(9,4)$