Question
If one of the diameters of the circle, given by the equation, $x^{2}+y^{2}-4 x+6 y-12=0$, is a chord of a circle $S$, whose centre is at $(-3,2)$, then the radius of $S$ is(A) 10(B) $5 \sqrt{2}$(C) $5 \sqrt{3}$(D) 5
Step 1
Comparing this with the standard equation, we find that the center of the circle is at $(2,-3)$ and the radius is $5$. Show more…
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If one of the diameters of the circle, given by the equation, $x^{2}+y^{2}-$ $4 x+6 y-12=0$, is a chord of a circle $S$, whose centre is at $(-3,2)$ then the radius of $S$ is: (a) 5 (b) 10 (c) $5 \sqrt{2}$ (d) $5 \sqrt{3}$
If one of the diameters of the circle $x^{2}+y^{2}-2 x-6 y+6=0$ is a chord to the circle with centre $(2,1)$, then the radius of the circle is (a) $\sqrt{3}$ (b) $\sqrt{2}$ (c) 3 (d) 2
If the length of the chord of the circle, $x^{2}+y^{2}=r^{2}(r>0)$ along the line, $y-2 x=3$ is $r$, then $r^{2}$ is equal to : (a) $\frac{9}{5}$ (b) 12 (c) $\frac{24}{5}$ (d) $\frac{12}{5}$
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