Question

If the curve $y=1+\sqrt{4-x^2}$ and the line $y=(x-2)k+4$ has two distinct points of intersection then the range of $k$ is

• A. $[1,3]$
• B. $\left[\frac{5}{12},\infty \right)$
• C. $\left(\frac{5}{12},\frac{3}{4}\right]$
• D. $\left[\frac{5}{12},\frac{3}{4}\right)$

Answer 1

The correct option is C $\left(\frac{5}{12},\frac{3}{4}\right]$

Equation of the line is $y-4=k(x-2)$
So it's a line passing through point $(2,4)$ with variable slope $k$
The curve represents upper half of a circle with radius $2$ and center $(0,1)$ as shown in the figure.
So, for two solutions, the straight line cuts circle at two distinct points. The point of intersection can be any point between $A$ to $B$ (excluding $B$).
Coordinates of point $A$ are $(-2,1)$.
Hence, slope of PA = $\frac{3}{4}$
B is the point of tangency by the line passing through point $(2,4)$.
Hence, $\left|\frac{-2k+4-1}{\sqrt{k^2+1}}\right|=2$
Solving we get $k=\frac{5}{12}$
Hence, slope of PB = $\frac{5}{12}$
$\therefore k\in (\frac{5}{12},\frac{3}{4}]$