Question

Consider points A$(\sqrt{13},0)$ and B$(2\sqrt{13},0)$ lying on x-axis. These points are roated in an anticlockwise direction about the origin through an angle of ${\tan }^{-1}\left(\frac{2}{3}\right)$. Let the new position of A and B be A' and B' respectively. With A' as centre and radius $\frac{2\sqrt{13}}{3}$ a circle C, is drawn and with B' as a centre and radius $\frac{\sqrt{13}}{3}$ circle $C_2$ is drawn. Find radical axis of $C_1$ and $C_2$.

Answer 1

given $A(\sqrt{13},0),B(2\sqrt{13},0)$ points are rotated in anti- clockwise direction about the origin through the angle of ${\tan }^{-1}\left(2/3\right)$

$x^{\prime }=x\cos \theta +y\sin \theta$

$y^{\prime }=-x\sin \theta +y\cos \theta$

$A^{\prime }(3,-2)$ $B^{\prime }(6,-4)$

Equation of the circle with centre $A^{\prime }$ & radius $\frac{2\sqrt{13}}{3}$

$C_1\cong (x-3{)}^2+(y+2{)}^2={\left(\frac{2\sqrt{13}}{3}\right)}^2$

Equation of the circle with centre $B^{\prime }$ & radius $\frac{\sqrt{13}}{3}$

$C_2\cong (x-6{)}^2+(y+4{)}^2={\left(\frac{\sqrt{13}}{3}\right)}^2$

Radical Axis of $C_1$ and $C_2$ is

$C_1-C_2=0$

$(x-3{)}^2+(y+2{)}^2-(x-6{)}^2-(y+4{)}^2={\left(\frac{2\sqrt{13}}{3}\right)}^2-{\left(\frac{\sqrt{3}}{3}\right)}^2$

$(x-3-x+6)(x-3+x-6)+(y+2-y-4)(y+2+y+4)=\frac{13}{9}$

$3[2x-9]-2[2y+6]=\frac{13}{9}$

$6x-27-4y-12=\frac{13}{9}$

$6x-4y-39-\frac{13}{9}=0$

$6x-4y-13\times \frac{28}{9}=0$

$3x-2y-\frac{13\times 14}{9}=0$

$27x-18y-182=0$