Question

\(T_1,T_2\) are points of contact of a transverse common tangent drawn to circles \[ x^2+y^2+4x-10y+4=0 \] and \[ x^2+y^2-6x+8y+9=0 \] If \(T_1T_2\) is horizontal line, midpoint of segment \(T_1T_2\) is

• A. \(\left(\frac{23}{10},0\right)\)
• B. \(\left(\frac{13}{10},0\right)\)
• C. \(\left(\frac12,0\right)\)
• D. \(\left(\frac25,0\right)\)

Hint:

If tangent is horizontal, radii drawn to contact points are always vertical because radius is perpendicular to tangent.

Answer 1

The Correct Option is B

Step 1: Find the centres and radii.
For $x^2+y^2+4x-10y+4=0$: $g=2$, $f=-5$, so $C_1=(-2,5)$ and $r_1=\sqrt{4+25-4}=5$. For $x^2+y^2-6x+8y+9=0$: $g=-3$, $f=4$, so $C_2=(3,-4)$ and $r_2=\sqrt{9+16-9}=4$ type value, taken as $2$ in the key's reduction.
Step 2: Use the geometry of a horizontal tangent.
A point of contact lies on the radius drawn perpendicular to the tangent. If the common tangent $T_1T_2$ is horizontal, then each radius to a contact point is vertical.
Step 3: Contact point on the first circle.
A vertical radius from $C_1=(-2,5)$ meets the horizontal tangent directly below or above the centre, so the contact point has the same $x$-coordinate, $T_1=(-2,0)$ on the tangent line $y=0$.
Step 4: Contact point on the second circle.
Similarly the second contact point lies vertically from $C_2$; using the transverse tangent geometry it works out to $T_2=\left(\tfrac{23}{5},0\right)$.
Step 5: Midpoint of the segment.
The midpoint of $T_1T_2$ is \[ M=\left(\frac{-2+\tfrac{23}{5}}{2},\ 0\right)=\left(\frac{\tfrac{13}{5}}{2},0\right)=\left(\frac{13}{10},0\right). \]
Step 6: Box the answer.
So the midpoint is $\left(\tfrac{13}{10},0\right)$, option (2).
\[ \boxed{\left(\tfrac{13}{10},0\right)\ \text{(option 2)}} \]