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
\(T_1,T_2\) 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
Show Hint
If tangent is horizontal, radii drawn to contact points are always vertical because radius is perpendicular to tangent.
Updated On: Jun 15, 2026
- \(\left(\frac{23}{10},0\right)\)
- \(\left(\frac{13}{10},0\right)\)
- \(\left(\frac12,0\right)\)
- \(\left(\frac25,0\right)\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Point of contact lies on radius perpendicular to tangent.
If common tangent is horizontal, radii to contact points are vertical.
Step 1: Find circle centers.
First circle: \[ C_1=(-2,5) \] Radius \[ r_1=5 \] Second circle \[ C_2=(3,-4) \] Radius \[ r_2=2 \]
Step 2: Horizontal tangent means contact points vertically aligned.
Thus contact points: \[ T_1=(-2,0) \] \[ T_2=\left(\frac{23}{5},0\right) \]
Step 3: Midpoint.
\[ M= \left( \frac{-2+\frac{23}{5}}{2},0 \right) \] \[ = \left( \frac{13}{10},0 \right) \] Hence \[ \boxed{\left(\frac{13}{10},0\right)} \]
Step 1: Find circle centers.
First circle: \[ C_1=(-2,5) \] Radius \[ r_1=5 \] Second circle \[ C_2=(3,-4) \] Radius \[ r_2=2 \]
Step 2: Horizontal tangent means contact points vertically aligned.
Thus contact points: \[ T_1=(-2,0) \] \[ T_2=\left(\frac{23}{5},0\right) \]
Step 3: Midpoint.
\[ M= \left( \frac{-2+\frac{23}{5}}{2},0 \right) \] \[ = \left( \frac{13}{10},0 \right) \] Hence \[ \boxed{\left(\frac{13}{10},0\right)} \]
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