TS EAMCET · Maths · Circle
Two sides of a square are along the lines \(x=-5\) and \(y=4\). The point of intersection of the diagonals is \((3,-4)\). The point of intersection of the tangents drawn to the circumcircle of the square at the two consecutive vertices lying on \(x=-5\) is
- A \((-4,-4)\)
- B \((-13,-4)\)
- C \((-4,-13)\)
- D \((-4,-10)\)
Answer & Solution
Correct Answer
(B) \((-13,-4)\)
Step-by-step Solution
Detailed explanation
For point \(\mathrm{R}(\mathrm{x}, \mathrm{y})\) \[ \begin{aligned} & \left(\frac{\mathrm{x}-5}{2}, \frac{\mathrm{y}+4}{2}\right)=(3,-4) \\ & \Rightarrow \mathrm{R}(11,-12) \end{aligned} \] Since PQRS is square \(\therefore \mathrm{Q}(-5,-12)\) and \(\mathrm{S}(4,11)\) Also PMQC…
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