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Question

Which of the following is a point on the common chord of the circle x2+y2+2x-3y+6=0 and x2+y2+x-8y-13=0?


A

(1,-2)

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B

(1,4)

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C

(1,2)

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D

(1,-4)

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Solution

The correct option is D

(1,-4)


Explanation for the correct option

Step 1: Finding the chord equation

The circle equations are,

x2+y2+2x-3y+6=0 and x2+y2+x-8y-13=0

If two circles are X and Y then the common cord of these two circles is X-Y=0

Here X=x2+y2+2x-3y+6 and Y=x2+y2+x-8y-13

x2+y2+2x-3y+6-(x2+y2+x-8y-13)=0

x2+y2+2x-3y+6-x2-y2-x+8y+13=0x+5y+19=0

Step 2: Substituting (1,-4) point on the L.H.S. of the equation x+5y+19=0 and comparing with R.H.S.

Substituting (1,-4) in the equation x+5y+19=0

(1)+5(-4)+19=1-20+19=0=R.H.S.

(1,-4) satisfies the chord equation.

Explanation for incorrect options

For option (A)

Substituting (1,-2) in the equation x+5y+19=0

(1)+5(-2)+19=1-10+19=10R.H.S.

For option (B)

Substituting (1,4) in the equation x+5y+19=0

(1)+5(4)+19=1+20+19=40R.H.S.

For option (C)

Substituting (1,2) in the equation x+5y+19=0

(1)+5(2)+19=1+10+19=30R.H.S.
Hence (1,-4) is the only point on the common chord of the circle x2+y2+2x-3y+6=0 and x2+y2+x-8y-13=0.

Therefore option (D) is the correct option.


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