Question

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

• A. $(1,-2)$
• B. $(1,4)$
• C. $(1,2)$
• D. $(1,-4)$

Answer 1

The correct option is D

$(1,-4)$

Explanation for the correct option

Step 1: Finding the chord equation

The circle equations are,

$x^2+y^2+2x-3y+6=0$ and $x^2+y^2+x-8y-13=0$

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

Here $X=x^2+y^2+2x-3y+6$ and $Y=x^2+y^2+x-8y-13$

$\Rightarrow x^2+y^2+2x-3y+6-(x^2+y^2+x-8y-13)=0$

$$\begin{gathered} \Rightarrow x^2+y^2+2x-3y+6-x^2-y^2-x+8y+13=0 \\ \Rightarrow x+5y+19=0 \end{gathered}$$

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

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

$$\begin{gathered} \left(1\right)+5(-4)+19 \\ =1-20+19 \\ =0 \\ =R.H.S. \end{gathered}$$

$(1,-4)$ satisfies the chord equation.

Explanation for incorrect options

For option (A)

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

$$\begin{gathered} \left(1\right)+5(-2)+19 \\ =1-10+19 \\ =10 \\ \ne R.H.S. \end{gathered}$$

For option (B)

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

$$\begin{gathered} \left(1\right)+5\left(4\right)+19 \\ =1+20+19 \\ =40 \\ \ne R.H.S. \end{gathered}$$

For option (C)

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

$$\begin{gathered} \left(1\right)+5\left(2\right)+19 \\ =1+10+19 \\ =30 \\ \ne R.H.S. \end{gathered}$$
Hence $(1,-4)$ is the only point on the common chord of the circle $x^2+y^2+2x-3y+6=0$ and $x^2+y^2+x-8y-13=0$.

Therefore option (D) is the correct option.