Question

The value of $\lambda$ for which the set $\left\{\left(x,y\right):x^2+y^2-6x+4y\le 12\right\}\cap$ $\left\{\left(x,y\right):4x+3y\le \lambda \right\}$ 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 $4x+3y\le \lambda$ must not satisfy the centre of the circle
$\therefore \lambda =-19$