Question

Find the radius of the circle which passes through the origin, $(0,4)$ and $(4,0)$.

• A. $\sqrt{8}$
• B. $4$
• C. $16$
• D. $\sqrt{36}$

Answer 1

The correct option is A $\sqrt{8}$

Circle passes through (0,0) , (0,4) & (4,0)
Equation of circle${⟹(x-a)}^2+{(y-b)}^2={\left(r\right)}^2$
At point $(0,0)$: $a^2+b^2=r^2⟶1$
At point $(0,4)$: $a^2+{(4-b)}^2=r^2⟶2$
At point $(4,0)$: ${(4-a)}^2+b^2=r^2⟶3$
from Equation 2 & 3
$a^2+{(4-b)}^2={(4-a)}^2+b^2$
$a^2+16+b^2-8b=16+a^2-8a+b^2$
$-8b=-8a$
$a=b$
from Equation 1 & 2
$a^2+b^2=a^2+{(4-b)}^2$
$b^2=16+b^2-8b$
$8b=16$
$b=2$
$\therefore a=2$
Put value of a and b in equation 1
$r^2=2^2+2^2=8$
$r=\sqrt{8}$