Question

The value of $\lambda$ for which the set $\left\{\right(x,y):x^2+y^2-6x+4y\le 12\}\cap \left\{\right(x,y):4x+3y\le \lambda \}$ contains only one point is

• A. $-31$
• B. $31$
• C. $-19$
• D. $19$

Answer 1

The correct option is C $-19$

$x^2+y^2-6x+4y-12=0$
Centre of the circle is $(3,-2)$ and radius is $\sqrt{9+4+12}=5$
Clearly, $4x+3y=\lambda$ should be tangent to the circle $x^2+y^2-6x+4y-12=0.$
$\therefore \left|\frac{4\left(3\right)+3(-2)-\lambda }{\sqrt{4^2+3^2}}\right|=5$
$\Rightarrow \lambda =31$ (or) $\lambda =-19$

Now, for only one point of intersection,
$4x+3y\le \lambda$ must not satisfy the centre of the circle.
$\therefore \lambda =-19$