Which of the following is a point on the common chord of the circles \( \Large x^{2}+y^{2}+2x-3y+6=0\ and\ x^{2}+y^{2}+x-8y-13=0 \)?


A) \( \Large \left(1,\ -2 \right) \)

B) \( \Large \left(1,\ 4\right) \)

C) \( \Large 1,\ 2 \)

D) \( \Large \left(1,\ -4\right) \)

Correct Answer:
D) \( \Large \left(1,\ -4\right) \)

Description for Correct answer:
Let the equation of circles are

\( \Large S_{1} = x^{2}+y^{2}+2x-3y+6=0 \) ...(i)

\( \Large S_{2} = x^{2}+y^{2}+x-8y-13=0 \) ...(ii)

Equation of common chord is

\( \Large S_{1} - S_{2} = 0 \)

=> \( \Large \left(x^{2}+y^{2}+2x-3y+6\right) - \left(x^{2}+y^{2}+x-8y-13\right) = 0 \)

=> \( \Large x+5y+19=0 \) ...(iii)

In the given options only the point \( \Large \left(1,\ -4\right) \) satisfied the eq. (iii)

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