Question

A tangent is drawn to the circle $2x^2+2y^2-3x+4y=0$ at the point 'A' and it meets the line $x+y=3$ at B(2, 1), then AB=______

• A. $\sqrt{10}$
• B. 0
• C. $2\sqrt{10}$
• D. 2

Answer 1

The correct option is B 0

Center of the circle is $(\frac{3}{4},1)$

Distance of point B from the center of the circle is $2-\frac{3}{4}=\frac{5}{4}$

The radius of the circle is given by $\sqrt{{\left(\frac{3}{4}\right)}^2+(1{)}^2}=\sqrt{\left(\frac{5}{4}\right)}$

Since distance between point B and the center of the circle equals the radius and a tangent touches the circle at only one point, A has to coincide with B.

$\therefore AB=0$