IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

Indian_Dragon

Find the radius of a circle having centre at (2,1) whose one of the chord is a diametre of the circle x² + y² -2x -6y +6 = 0

fighter

centre of the circle x^2+y^2-2x-6y+6 is (1,3)

half of the length of chord is 2 unit.

distance of point from(2,1) =root5units

radius of the circle will be root(5+2^2) i.e. 3 units

eq of circle will be

(x-2)^2 + (y-1)^2=9

raulrag009

Here

Eqn of circle will be

(x-2)²+(y-1)²=r²

x²+y²-4x-2y+5-r²=0

And the other circle is

x²+y²-2x-6y+6=0

Eqn of common chord {subtract both the above eqn's}

2x-4y+1+r²=0

x-2y + (1+r²)/2 =0..........(1)

Now,

Eqn of chord bisected at a point{1,3,centre of other circle} is given by

T=S₁

x(1)+y(3)-2(x+1)-(y+3)+5-r²=1+9-4-6+5-r²

x-2y+5=0................(2)

comparing (1)and (2)

we get

(1+r²)/2 = 5

1+r²=10

r²=9

r=+-3

thus radius is 3 units