Question

A square is formed by following two pairs of straight lines $y^2-14y+45=0$ and $x^2-8x+12=0$. A circle is inscribed in it. Then the centre of the circle is

• A. $(7,4)$
• B. $(4,7)$
• C. $(3,7)$
• D. $(\frac{3}{8},4)$

Answer 1

Answer: (B)

$(4,7)$

$y^2-14y+45=0$

$y=\frac{14\pm \sqrt{196-180}}{2}$

$y=\frac{14\pm 4}{2}$

$y=9,5$

and $x^2-8x+12=0$

$x=\frac{8\pm \sqrt{64-48}}{2}$

$x=6,2$

side of square $ABCD$ is $4$

so , radius of circle $=r=\frac{4}{2}=2$

and center of the circle is the midpoint of diagonal RD

$\Rightarrow center(4,7)$