Question

Find the interval of \(\lambda\) for which exactly two common tangents can be drawn to \(x^2+y^2-4x-4y+6=0\) and \(x^2+y^2-10x-10y+\lambda=0\).

• A. (12, 24)
• B. (12, 32)
• C. (18, 42)
• D. (18, 48)

Hint:

Always verify that the radius remains a real number (\(\lambda < 50\)) when finding the interval for \(\lambda\).

Answer 1

The Correct Option is C

Step 1: Recall the two-tangent condition.
Two circles have exactly two common tangents precisely when they intersect at two points, i.e. $|r_1-r_2|<d<r_1+r_2$, where $d$ is the distance between centres.
Step 2: Find circle 1's centre and radius.
For $x^2+y^2-4x-4y+6=0$: centre $(2,2)$ and $r_1^2=2^2+2^2-6=2$, so $r_1=\sqrt2$.
Step 3: Find circle 2's centre and radius.
For $x^2+y^2-10x-10y+\lambda=0$: centre $(5,5)$ and $r_2=\sqrt{50-\lambda}$.
Step 4: Compute the distance between centres.
\[ d=\sqrt{(5-2)^2+(5-2)^2}=\sqrt{18}=3\sqrt2. \]
Step 5: Apply the upper inequality $d<r_1+r_2$.
$3\sqrt2<\sqrt{50-\lambda}+\sqrt2$ gives $2\sqrt2<\sqrt{50-\lambda}$, so $8<50-\lambda$, hence $\lambda<42$.
Step 6: Apply the lower inequality $d>|r_1-r_2|$.
$3\sqrt2>\sqrt{50-\lambda}-\sqrt2$ gives $4\sqrt2>\sqrt{50-\lambda}$, so $32>50-\lambda$, hence $\lambda>18$. Combining, $\lambda\in(18,42)$, which is option 3.
\[ \boxed{\lambda\in(18,42)} \]