AP EAMCET · Maths · Circle
The power of the point \(B(-1,1)\) with respect to the circle \(S \equiv x^2+y^2-2 x-4 y+3=0\) is \(p\). If the length of the tangent drawn from \(B\) to the circles \(S=0\) is \(t\), then the point \((2,3)\) with respect to the circle \(S^{\prime}=0\) having centre at \(\left(p, t^2\right)\) and passing through the origin.
- A lies inside the circle S' = 0
- B lies outside the circle S' = 0
- C lies on the circle S' = 0
- D is the centre of the circle S' = 0
Answer & Solution
Correct Answer
(A) lies inside the circle S' = 0
Step-by-step Solution
Detailed explanation
Given equation of circle \[ \begin{array}{rlrl} & & S & \equiv x^2+y^2-2 x-4 y+3=0 \\ \therefore & \quad p & =(-1)^2+(1)^2-2(-1)-4(1)+3 \\ & & =1+1+2-4+3=3 \\ \because & & t & =\sqrt{p} \Rightarrow t=\sqrt{3} \end{array} \] Now, circle whose centre is \(\left(p, t^2\right)\),…
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