Question

Let P, Q, R and S be the feet of the perpendiculars drawn from a point (1, 1) upon the lines x + 4y = 12, 4y - x = 4 and their angle bisectors respectively then equation of the circle which passes through Q, R, S is

• A. $x^2+y^2-5x-3y+6=0$
• B. $x^2+y^2+104x-110+0$
• C. $3x^2+3y^2-4x-18y+16=0$
• D. none of these

Answer 1

Answer: (A)

$x^2+y^2-5x-3y+6=0$

Given lines

$x+4y=12$----(1)

$-x+4y=4$----(2)

Equation of angle bisector of line (1) and (2)

$x+4y-12=\pm (-x+4y-4)$

$x+4y-12=-x+4y-4$ and $x+4y-12=+x-4y+4$

$x=4$----(3) and $y=2$----(4)

Point P is the foot of perpendicular on line $x+4y-12=0$ from $(1,1)$

BY foot of perpendicular formula

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\left(\frac{a{x}_1+b{y}_1+c}{a^2+b^2}\right)$

$\frac{x-1}{1}=\frac{y-1}{4}=-\left(\frac{1+4-12}{1^2+4^2}\right)$

$\frac{x-1}{1}=\frac{y-1}{4}=-\left(\frac{-7}{17}\right)$

$\frac{x-1}{1}=\frac{y-1}{4}=\frac{7}{17}$

$\frac{x-1}{1}=\frac{7}{17}$

$x=\frac{7}{17}+1\Rightarrow x=\frac{24}{17}$

Similarly $y=\frac{45}{17}$

$P(\frac{24}{17},\frac{45}{17})$

Point Q is the foot of perpendicular on line $-x+4y-4=0$ from $(1,1)$

BY foot of perpendicular formula

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\left(\frac{a{x}_1+b{y}_1+c}{a^2+b^2}\right)$

$\frac{x-1}{-1}=\frac{y-1}{4}=-\left(\frac{-1+4-4}{(-1{)}^2+4^2}\right)$

$\frac{x-1}{-1}=\frac{y-1}{4}=-\left(\frac{-1}{17}\right)$

$\frac{x-1}{-1}=\frac{y-1}{4}=\frac{1}{17}$

$\frac{x-1}{-1}=\frac{1}{17}$

$x=\frac{-1}{17}+1\Rightarrow x=\frac{16}{17}$

Similarly $y=\frac{21}{17}$

$Q(\frac{16}{17},\frac{21}{17})$

Point R is the foot of perpendicular on line $x-4=0$ from $(1,1)$

BY foot of perpendicular formula

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\left(\frac{a{x}_1+b{y}_1+c}{a^2+b^2}\right)$

$\frac{x-1}{1}=\frac{y-1}{0}=-\left(\frac{1-4}{1^2}\right)$

$\frac{x-1}{1}=\frac{y-1}{0}=-\left(\frac{-3}{1}\right)$

$\frac{x-1}{1}=\frac{y-1}{0}=3$

$\frac{x-1}{1}=3$

$x=3+1\Rightarrow x=4$

Similarly $y=1$

$R(4,1)$

Point S is the foot of perpendicular on line $y-2=0$ from $(1,1)$

BY foot of perpendicular formula

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=-\left(\frac{a{x}_1+b{y}_1+c}{a^2+b^2}\right)$

$\frac{x-1}{0}=\frac{y-1}{1}=-\left(\frac{1-2}{1^2}\right)$

$\frac{x-1}{0}=\frac{y-1}{1}=-\left(\frac{-1}{1}\right)$

$\frac{x-1}{0}=\frac{y-1}{1}=1$

$\frac{x-1}{0}=3$

$x=1$

Similarly $y=2$

$S(1,2)$

Here eq of circle passing through $Q,R,S$

and Here line $QR$ and $QS$ is perpendicular So eq of circle from R and S by formula

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

$(x-4)(x-1)+(y-1)(y-2)=0$

$x^2+4-5x+y^2+2-3y=0$

$x^2+y^2-5x-3y+6=0$