Question

The radius of a circle with center $\left(a,b\right)$ and passing through the center of the circle $x^2+y^2-2gx+f^2=0$ is:

• A. $\sqrt{{\left(a-g\right)}^2+b^2}$
• B. $\sqrt{a^2+{\left(b+g\right)}^2}$
• C. $\sqrt{a^2+{\left(b-g\right)}^2}$
• D. $\sqrt{{\left(a+g\right)}^2+b^2}$

Answer 1

The correct option is A $\sqrt{{\left(a-g\right)}^2+b^2}$

$x^2+y^{2}2gx+f^2=0$
Center $(a,b)$
The correct equation are possible from circle
$x^2+y^2-2gx+g^2=0⟶\left(1\right)x^2+y^2-2gx=f^2⟶\left(2\right)Testing\left(1\right)x^2+y^2-2gx+g^2=0x^2-2gx+g^2+y^2=0{(x-g)}^2+{(y-0)}^2=0$
Thus the center here of circle is $(g,0).$
As circle A passes brought this point.
$x^2+y^2-2gx=f^2{x}^2+y^2-2gx-f^2=0$
Equation of circle A.
${(x-g)}^2+{(y-f)}^2=r^2$
When the center lies on $x-axis$, then $y$ we have will be zero.
Taking $f=0{(x-g)}^2+{(y-f)}^2={(x-g)}^2+{(y-0)}^2={(x-g)}^2+y^2=r^2{x}^2+g^2-2gx+y^2=r^2$
Putting $r^2=g^2+f^2{x}^2+y^2-2gx+g^2=g^2+f^2{x}^2-2gx+y^2=f^2$
$\therefore$ Point $(a,b)(g,0)r=\sqrt{(a-g{)}^2}+b^2$