# Finding number of circles and area of polygon formed by their centres.

Let $S(x,y)=0$ represent a circle with radius $\frac{3}{\sqrt 2}$ such that $S(\lambda -3, \lambda) =0$ has equal roots and $S(\mu, 7- \mu )=0$ also has equal roots then find

1) number of such circles

2) number of circles whose centre lie in first quadrant

3) area of polygon formed by joining all possible centres.

My approach: I wrote $S=0$ in the general form of circle with unknowns g, f, and c. Using the condition of radius I got one equation in these variables. Then substituted the given points in equation of circle to get quadratic equations in $\lambda$ and $\mu$. Then I set the discriminant of these equations equal to 0 to obtain two more equations. But the equations obtained appear to be impossible to solve because of so many of variables in 2nd degree. Can anyone please provide me some hints to deal with this question in an easy way.

• What do you mean by equal roots? – mephistolotl Nov 26 '17 at 5:58
• On substituting ($\lambda-3, \lambda$) in the equation of unknown circle in place of (x, y) you get two equal roots of $\lambda$ i.e. there is only one solution to the quadratic of $\lambda$ thus obtained – Rohan Shinde Nov 26 '17 at 6:07
• Ah, I'm used to the terms "double root" and "repeated root". Just out of curiousity, where are you from? I'm from western Canada. – mephistolotl Nov 26 '17 at 7:44

The following is just painful computation from where you left off. I would be very interested to see if there is a slicker way to do this problem; in particular, why did the magic cancelations happen?

The general equation is $$S(x,y) = (x-a)^2 + (y-b)^2 - 9/4.$$

Then \begin{align} S(\lambda - 3, \lambda) &= (\lambda - (a+3))^2 + (\lambda - b)^2 - 9/4 = 2 \lambda^2 - 2(a+b+3)\lambda + (a+3)^2 + b^2 - 9/4 \\ S(\mu, 7-\mu) &= (\mu - a)^2 + (7-\mu-b)^2 - 9/4 = 2\mu^2 -2(a-b+7)\mu + a^2 + (b-7)^2 - 9/4 \end{align}

Setting the discriminants to zero yields \begin{align} (a+b+3)^2 - 2(a+3)^2 - 2b^2 + 9/4=0 \\ (a-b+7)^2 - 2a^2 - 2(b-7)^2 + 9/4=0 \end{align}

If we write the left-hand sides as quadratics in $b$, then we have \begin{align} b^2 -(2a+6)b+(a^2 +6a+27/4) &= 0\\ b^2 +(2a-14)b+(a^2 -14a+187/4) &= 0 \end{align}

If you use the quadratic formula to solve for $b$, you get magic cancelations that lead to \begin{align} b &= \frac{1}{2} [2a+6 \pm \sqrt{(2a+6)^2-4(a^2 + 6a+27/4)}] = a + 3 \pm 3/2 \\ b &= \frac{1}{2}[-2a + 14 \pm \sqrt{(2a-14)^2 + 4(a^2 - 14a + 187/4)}] = -a +7 \pm 3/2 \end{align}

Thus, the solutions to "$S(\lambda-3,\lambda)$ has repeated roots" are the parallel lines $b=a+9/2$ and $b=a+3/2$, while the solutions to "$S(\mu, 7-\mu)$ has repeated roots" are the parallel lines $b=-a+11/2$ and $b=-a+17/2$.

Thus, there are four solutions, and it should be easy to finish the problem from here.

• So, will there be exactly 4 circles due to the intersection of the non-parallel lines? – Your IDE Nov 26 '17 at 6:48
• @YourIDE Yes, that is what I tried to say at the end of my post. – angryavian Nov 27 '17 at 2:22

Hint: the two given conditions that is $S(\lambda-3, \lambda)=0$ and $S(\mu, 7-\mu)=0$ can be interpreted geometrically as two tangents with equations $y-x=3$ and $x+y=7$ respectively to the circle $S=0$ of radius $\frac{3}{\sqrt2}$. Thus given questions can be answered on given basis. Therefore the centres are (5,5), (2,2), (2,8), (-1,5)