Question

Consider a family of circles, each passing through (1, 1) and having its center on the bisector of the acute angle between the lines $\sqrt{3}(y-1)=x-1$ and $y-1=\sqrt{3}(x-1)$. Prove that the common chords of each member of the family and circle $x_2+y_2+4x-4y-1=0$ are concurrent. Find the point of concurrency.

Answer 1

On interchanging $x$ and $y$ of$\sqrt{3}(y-1)=(x-1)$ we get,

$(y-1)=\sqrt{3}(x-1)$ which is the second equation.

ie, both equations are mirror images of each other on $y=x$, which is therefore the angle bisector of both lines.

Or, you can see that Slope of 1st equation, $\tan {\theta }_1=\frac{1}{\sqrt{3}}$.

$\Rightarrow {\theta }_1={30}^0$

and the slope of 2nd equation, $\tan {\theta }_2=\sqrt{3}$

$\Rightarrow {\theta }_2={60}^0$

So, Acute angle$={60}^0-{30}^0={30}^0$

So, the angle which can bisect the angle will be, ${30}^0+\frac{{30}^0}{2}={45}^0$ (Half of the acute angle on top of lowest angle)

So, the equation of the bisector will be, $(y-1)=\tan \left({45}^0\right)(x-1)\Rightarrow y=x$

So, the family of circles with centres on it will have centres $(h,h)$, such that $h\in$Real Numbers.

So, the family of circles which goes through $(1,1)$ and have centres $(h,h)$ are,

$(x-h{)}^2+(y-h{)}^2=(1-h{)}^2+(1-h{)}^2=2(1-h{)}^2\Rightarrow x^2+y^2-2xh-2yh=2-4h$

So, the equation of family of common chords of these circles will have with $x^2+y^2+4x-4y=1$ is,

$(x^2+y^2+4x-4y)-(x^2+y^2-2xh-2yh)=1-(2-4h)\Rightarrow (4+2h)x-(4-2h)y=4h-1$

As all the family of lines pass through a same point, the equation should be independent of $h$ at that point.

ie, $2hx+2hy=4h\Rightarrow x+y=2$

and $4x-4y=(-1)\Rightarrow x-y=\frac{(-1)}{4}$

From adding and substracting both equations, $(x,y)=(\frac{7}{8},\frac{9}{8})$ which is the common point through which all the chords pass through.