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MHT CET · Maths · Circle

The equation of common tangent to the circles \(\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}+10 \mathrm{y}+20=0\) and \(\mathrm{x}^2+\mathrm{y}^2+8 \mathrm{x}-6 \mathrm{y}-24=0\) is

  1. A \(3 x-4 y+11=0\)
  2. B \(3 x-4 y-11=0\)
  3. C \(-3 x-4 y+11=0\)
  4. D \(3 x+4 y+11=0\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3 x-4 y-11=0\)

Step-by-step Solution

Detailed explanation

Circle \(x^2+y^2-4 x+10 y+20=0\) has centre \(C_1=(2,-5)\) and radius
\(
\mathrm{r}_1=\sqrt{4+25-20}=3
\)
Circle \(\mathrm{x}^2+\mathrm{y}^2+8 \mathrm{x}-6 \mathrm{y}-24=0\) has centre \(\mathrm{C}_2=(2,-5)\) and radius
\(
\mathrm{r}_2=\sqrt{16+9+24}=7
\)
Distance between centres
\(
=\sqrt{(2+4)^2+(5-2)^2}=10
\)
Thus circle touch each other externally at one point only.
\(\therefore\) Equation of common tangent is
\(\left(x^2+y^2-4 x+10 y+20\right)-(x^2+y^2+8 x-6 y\)\(-24)=0 \text { i.e. } \)
\( 12 x-16 y-44=0 \quad \Rightarrow \quad 3 x-4 y-11=0 \)