AP EAMCET · Maths · Circle
The point at which the circles \(x^2+y^2-4 x-4 y+7=0 \quad\) and \(x^2+y^2-12 x\) \(-10 y+45=0\) touch each other, is
- A \(\left(\frac{13}{5}, \frac{14}{5}\right)\)
- B \(\left(\frac{2}{5}, \frac{5}{6}\right)\)
- C \(\left(\frac{14}{5}, \frac{13}{5}\right)\)
- D \(\left(\frac{12}{5}, 2+\frac{\sqrt{21}}{5}\right)\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{14}{5}, \frac{13}{5}\right)\)
Step-by-step Solution
Detailed explanation
Centres and radii of given circles are \[ C_1(2,2), r_1=\sqrt{2^2+2^2}-7=1 \] and \(C_2(6,5)\), \[ \begin{aligned} r_2 & =\sqrt{5^2+5^2-45} \\ & =\sqrt{36+25-45}=4 \end{aligned} \] Let \(P\) be the point at which the circle touch. Using internal ratio formula,…
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