Question

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

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

Answer 1

Answer: (D)

Given line $L:4x+3y=\lambda $
Circle $S:x^{2}+y^{2}-6x+4y=12$
Centre $\left(3 , - 2\right),$ Radius $=5$
Now, the line will be tangent to the circle
$\therefore $ $p=r$
$\left|\frac{4 \left(3\right) + 3 \left(- 2\right) - \lambda }{5}\right|=5\Rightarrow \lambda =31,-19$
So, for only one point of intersection, inequality $4x+3y\leq \lambda $ must not satisfy the centre of the circle
$\therefore \lambda =-19$