Question

If one of the diameters of the circle, given by the equation $x^2+y^2-4x+6y-12=0$, is a chord of a circle S, whose centre is at (-3,2), then the radius of S is

• A. $5\sqrt{2}$
• B. $5\sqrt{3}$
• C. $5$
• D. $10$

Answer 1

The correct option is B $5\sqrt{3}$

Given equation of a circle is $x^2+y^2-4x+6y-12=0$
Whose centre is (2, -3) and radius
$=\sqrt{2^2+(-3{)}^2+1}2=\sqrt{4+9+12}=5$



Now, according to given information, we have the following figure.
$x^2+y^2-4x+6y-12=0$
Clearly, $AO\perp BC$ ad O is mid-point of the chord.
Now, in $\Delta AOB$, we have
$OA=\sqrt{(-3-2{)}^2+(2+3{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$
And OB=5
$\therefore AB=\sqrt{O{A}^2+O{B}^2}=\sqrt{50+25}=\sqrt{75}=5\sqrt{3}$