Question

Sides of triangle are $3x-4y=5,4x+3y=6$ and $x+y=7$. Taking these sides as diameter, if we draw three circles, then the radical centre of these three circles is

• A. $(\frac{39}{5},\frac{-2}{5})$
• B. $(\frac{39}{25},\frac{-2}{25})$
• C. $(\frac{34}{5},\frac{-1}{5})$
• D. $(\frac{34}{25},\frac{-2}{25})$

Answer 1

The correct option is B $(\frac{39}{25},\frac{-2}{25})$

Let the 3 lines meet at A, B and C then radical center will be the orthocenter of triangle ABC.

$m_{e_1}\times m_{e_2}=\frac{3}{4}\times \frac{-4}{3}=-1$

$⟹l_1\perp l_2⟹$ ABC is right triangle $⟹$ vertex at right angle (A) is orthocenter.

$l_1=3x–4y=5$

$l_2=4x+3y=6$

On solving the $l_1$ and $l_2$,

$x=\frac{39}{25},y=\frac{-2}{25}$

$RC=(\frac{39}{25},\frac{-2}{25})$