Question

A pair of straight lines $x^2-8x+12=0$ and $y^2-14y+45=0$ are forming a square. Co-ordinates of the center of the circle inscribed in the square are

• A. (3, 6)
• B. (4, 7)
• C. (4, 8)
• D. none

Answer 1

The correct option is B (4, 7)

Given pair of straight lines $x^2-8x+12=0$ and $y^2-14y+45=0$
Consider $x^2-8x+12=0$
$\Rightarrow x^2-6x-2x-12=0$
$\Rightarrow x(x-6)-2(x-6)=0$
$\Rightarrow (x-6)(x-2)=0$
$\therefore x=6,x=2$
Similar way for $y^2-14y+45=0$
$\Rightarrow y^2-9y-5y+45=0$
$\Rightarrow y(y-9)-5(y-9)=0$
$\Rightarrow (y-9)(y-5)=0$
$\therefore y=9,y=5$
Centre of the circle inscribed in the square will be $(\frac{2+6}{2},\frac{5+9}{2})$
$=(\frac{8}{2},\frac{14}{2})$
$=(4,7)$