Question

Find the equation of circle circumscribing the quadrilateral formed by four lines $3x+4y-5=0$, $4x-3y-5=0$, $3x+4y+5=0$ and $4x-3y+5=0$

• A. x² + y² = 0
• B. x² + y² = 2
• C. x² + y² = 25
• D. x² + y² = 4

Answer 1

The correct option is B

x² + y² = 2

Given four sides of quadrilateral are

$L_{1}3x+4y-5=0$

$L_{2}4x-3y-5=0$

$L_{3}3x+4y+5=0$

$L_{4}4x-3y+5=0$

Equation of the circle circumscribing a quadrilateral whose sides in order are represented by $L_1=0,L_2=0,L_3=0$ and $L_4=0$ is

$\mu L_1{L}_3+\lambda L_2{L}_4=0$

$\mu (3x+4y-5)(3x+4y+5)+\lambda (4x-3y-5)(4x-3y+5)=0$

$\mu \left(\right(3x+4y{)}^2-25)+\lambda ((4x-3y{)}^2-25)=0$

$\mu (9x^2+16y^2+24xy-25)+\lambda (16x^2+9y^2-24xy-25)=0$

$(9\mu +16\lambda )x^2+(16\mu +9\lambda )y^2+(24\mu -24\lambda )xy-(25\mu +25\lambda )=0$- - - - - - (1)

Since, this is an equation of circle

Conditions for circle

(1) Coefficient of xy = 0

$24\mu -24\lambda =0$

$\mu =\lambda$

(2) Coefficient of $x^2=$coefficient of $y^2$

$9\mu +16\lambda =16\mu +9\lambda$

$-7\mu +7\lambda =0$

$\lambda =\mu$

so, given equation of circle is true for any value of $\lambda =\mu$ except $\lambda =\mu \ne 0$

lets $\lambda =\mu =1$

Substituting the value of $\lambda =\mu =1$ in equation (1)

$(9+16)x^2+(16+9)y^2+0\times xy-(25+25)=0$

$25x^2+25y^2-50=0$

$x^2+y^2=2$

Equation of circle circumscribing the quadrilateral formed by lines $3x+4y-5=0,4x-3y-5=0,3x+4y+5=0,4x-3y+5=0$ is

$x^2+y^2=2$

Option B is correct