Question

For hyperbola $\frac{x^2}{16}-\frac{y^2}{12}=1$, and if three concyclic points are $(4.5,1.9),(-4.5,1.9),(-4.5,-1.9)$. the fourth concyclic point and equation of circle is

• A. $(4.5,-2)$, $x^2+y^2=36$
• B. $(-4.5,-3)$, $x^2+y^2=49$
• C. $(4.5,-1.9)$, $x^2+y^2=25$
• D. $(4.5,-2.9)$, $x^2+y^2=16$

Answer 1

The correct option is C $(4.5,-1.9)$, $x^2+y^2=25$

Given : $\frac{x^2}{16}-\frac{y^2}{16}=1$

Three concyclic points $(4.5,1.9),(-4.5,1.9),(-4.5,-1.9)$

All these points are vertices of rectangle

Fourth vertex is $(4.5,-1.9)$

$r=\sqrt{{4.5}^2+{1.9}^2}\approx 5$

$x^2+y^2=25$