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Question

Mathematics Question on Conic sections

Equation of chord of the circle x2+y2+4x−6y−9=0x^2 + y^2 + 4x - 6y - 9 = 0 bisected at (0, 1) is

A

y−1=x y - 1 = x

B

y+1=x y + 1 = x

C

y+1=2x y + 1 = 2x

D

y−1=3x y - 1 = 3x

Answer

y−1=x y - 1 = x

Explanation

Solution

Since chord of circle
x2+y2+4x−6y−9=0x^2 + y^2 +4x - 6y- 9 = 0 bisected at (0, 1)
⇒   OC⊥AB\Rightarrow \:\:\: OC \bot AB
∴\therefore Slope of OC×OC \times Slope of AB=−1AB= -1
Centre of given circle is (-2, 3) and mid-point of chord is (0,1)(0,1)
Let any other point of chord is (x,y) (x, y) then slope of chord is y−1x−0\frac{y-1}{x-0} and slope of OC=3−1−2−0OC = \frac{3-1}{-2-0}
∴   (3−1−2−0)(y−1x−0)=−1\therefore \:\:\: \left(\frac{3-1}{-2-0}\right) \left(\frac{y-1}{x-0}\right) = - 1
or (2−2)(y−1x)=−1\left(\frac{2}{-2}\right) \left( \frac{y-1}{x}\right) = -1
or y−1x=1\frac{y-1}{x} = 1 or y−1=xy - 1 =x
is the required equation of chord.