2

Question: If one of the diameters of the circle $x^2 + y^2-2x-6y+6 = 0$ is a chord to the circle with center (2, 1), then the radius of the circle is:

$\sqrt3,\sqrt2,3,2$

I have no clue as to where to begin this question. I tried drawing a diagram, but that didn't help me.

Gummy bears
  • 3,408
  • 1
    Do you want the radius of the circle $x^2+y^2-2x-6y+6 = 0$ or the radius of the circle centered at $(2,1)$? I'd have to guess the second option, but this is not clear from the wording of the question. – JimmyK4542 Aug 28 '14 at 16:43

1 Answers1

2

So, the radius of the given circle $=\sqrt{1^2+3^2-6}=2$

Using Perpendicular Bisector of Chord Passes Through Centre

Observe that the distance between the two centres , this radius($=2$) & the other radius(hypotenuse) form a Pythagorean triangle

lab bhattacharjee
  • 274,582