Question

If the equation $a{x}^2+2(a^2+ab-16)xy+b{y}^{2}2ax+2by-\sqrt[4]{42}=0$ represents a circle, the radius of the circle is

• A. $\sqrt{16+\sqrt[4]{42}}$.
• B. $\sqrt{24+\sqrt[4]{42}}$.
• C. $\sqrt{2}$
• D. $\sqrt[4]{42}$

Answer 1

Answer: (A)

$\sqrt{16+\sqrt[4]{42}}$.

$a{x}^2+2(a^2+ab-16)xy+b{y}^2+2ax+2by-\sqrt[4]{42}=0$

If the above equation represents the equation of circle then-

$a=b.....\left(1\right)$

$a^2+ab-16=0$

$\Rightarrow a^2+a^2-16=0$

$\Rightarrow 2a^2=16$

$\Rightarrow a^2=8$

$\therefore b^2=a^2=8$

Radius of circle $\left(r\right)=\sqrt{a^2+b^2-c}$

$\Rightarrow r=\sqrt{8+8+\sqrt[4]{42}}$

$\Rightarrow r=\sqrt{16+\sqrt[4]{42}}$

Hence the radius of the circle is $\sqrt{16+\sqrt[4]{42}}$.