Question

A parabola $y^2=4x$ and a line $x-y=1$ intersect at two points, taking these intersection points as ends of a diameter, a circle is drawn. The centre of the circle is

• A. $(4,3)$
• B. $(3,2)$
• C. $(\frac{3}{2},\frac{1}{2})$
• D. $(\frac{1}{2},\frac{-1}{2})$

Answer 1

The correct option is B $(3,2)$

$y^2=4x$................$\left(1\right)$

$x-y=1$.............$\left(2\right)$

$\Rightarrow y^2=4(1+4)$

$\begin{array}{c}\Rightarrow y^2-4y-4=0\\ \Rightarrow y=\frac{4\pm \sqrt{16+16}}{2}\\ =\frac{4\pm 4\sqrt{2}}{2}\\ \Rightarrow y=2\pm 2\sqrt{2}\\ ify=\left(2+2\sqrt{2}\right)thenx=\left(3+2\sqrt{2}\right)\\ soA=\left(3+2\sqrt{2},2-2\sqrt{2}\right)\\ centreofcircle\\ =\left(\frac{3+2\sqrt{2}+3-2\sqrt{2}}{2},\frac{2+2\sqrt{2}+2-2\sqrt{2}}{2}\right)\\ henceweget\\ =\left(3,2\right)\end{array}$

so option $B$ is correct answer.