Question

The circle ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$ and hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$. Equation of the circle with $AB$ as its diameter is

• A. $x^2+y^2-12x+24=0$
• B. ${\mathrm{x}}^2+{\mathrm{y}}^2+12\mathrm{x}+24=0$
• C. $x^2+y^2-24x-12=0$
• D. $x^2+y^2-24x+24=0$

Answer 1

The correct option is A

$x^2+y^2-12x+24=0$

Explanation for the correct option:

Finding the equation of the circle:

Given, the equation of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$and the equation of circle is ${\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}=0$

The points of intersection is $\frac{x^2}{9}+\frac{(x^2-8x)}{4}=1$

Simplifying the equation we get,

$$\begin{gathered} 4x^2+9x^2–72x=36 \\ 13x^2–72x–36=0 \\ x=6\mathrm{and}x=\frac{-13}{6} \end{gathered}$$

But $\mathrm{x}=\frac{-13}{6}$ is not acceptable as it is a negative value

$$\begin{gathered} \mathrm{x}=6 \\ \text{Substituting the value of}x\text{in} \\ \frac{x^2}{9}-\frac{y^2}{4}=1 \\ \frac{36}{9}-\frac{y^2}{4}=1 \\ 4-\frac{y^2}{4}=1 \\ \frac{y^2}{4}=3 \\ {y}^2=12 \\ \mathrm{y}=\pm 2\sqrt{3} \end{gathered}$$

Therefore, the required equation is

$$\begin{gathered} {(x–6)}^2+(y+2\sqrt{3})(y–2\sqrt{3})=0 \\ {x}^2+y^2–12x+24=0 \end{gathered}$$

Hence, option(A) is correct.