AP EAMCET · Maths · Circle
The distance between the centres of similitude of the circles \(x^2+y^2+6 x-8 y+16=0\) and \(x^2+y^2-2 x-2 y+\) \(1=0\) is
- A \(\frac{15}{4}\)
- B \(\frac{5}{4}\)
- C \(\frac{5}{2}\)
- D \(\frac{15}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{15}{4}\)
Step-by-step Solution
Detailed explanation
Given circles \(x^2+y^2+6 x-8 y+16=0\) and \(x^2+y^2-2 x-2 y+1=0\) \(\Rightarrow(x+3)^2+(y-4)^2=3^2,(x-1)^2+(y-1)^2=1^2\) So \(C_1=(-3,4), C_2=(1,1), r_1=3 \& r_2=1\) Let internal centre \(P=\left(\frac{3(1,1)+1(-3,4)}{1+3}\right)\) \(=\left(0, \frac{7}{4}\right)\) external…
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