JEE Advanced 2010 Paper 1Question 1
Equation of the circle with AB as its diameter is
Circle
Solution
Concept. From the paragraph: the circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}9-\frac{y^2}4=1$ meet at $A$ and $B$. Solve them together, then write the circle on $AB$ as diameter. Formulas used. (1) Circle on $(x_1,y_1),(x_2,y_2)$ as diameter: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.Step 1 - Eliminate $y^2$. $y^2=8x-x^2$ in $4x^2-9y^2=36$: $$4x^2-72x+9x^2=36\Rightarrow13x^2-72x-36=0\Rightarrow(x-6)(13x+6)=0.$$ Step 2 - Reject. $x=-\frac6{13}$ gives $y^2=8x-x^2<0$. So $x=6$, $y^2=48-36=12$. Step 3 - Points. $A=(6,2\sqrt3)$, $B=(6,-2\sqrt3)$.Step 4 - Use (1). $$(x-6)^2+(y-2\sqrt3)(y+2\sqrt3)=0\Rightarrow x^2+y^2-12x+24=0.$$ Answer: A