TS EAMCET · Maths · Circle
If the equation of the transverse common tangent of the circles \(x^2+y^2-4 x+6 y+4=0\) and \(x^2+y^2+2 x-2 y-2=0\) is \(a x+b y+c=0\), then \(\frac{a}{c}=\)
- A \(-\frac{3}{4}\)
- B \(\frac{4}{3}\)
- C 1
- D -1
Answer & Solution
Correct Answer
(D) -1
Step-by-step Solution
Detailed explanation
Centre and radius of given circles \(C_1(2,-3), r_1=3 ; C_2(-1,1), r_2=2\) \(\begin{aligned} & \because \mathrm{C}_1 \mathrm{C}_2=\sqrt{(2+1)^2+(1+3)^2}=5 \\ & r_1 r_2=3+2=5 \\ & \because \mathrm{C}_1 \mathrm{C}_2=\mathrm{n}_2 \end{aligned}\) So, both circles touch at \(P\).…
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