Let be the circle of radius 1 with center at the origin. Let be the circle of radius with center at the point A = (4, 1), where . Two distinct common tangents PQ and ST of and are drawn. [2023]

Question

Let $C_{1}$ be the circle of radius 1 with center at the origin. Let $C_{2}$ be the circle of radius $r$ with center at the point A = (4, 1), where $1 < r < 3$ . Two distinct common tangents PQ and ST of $C_{1}$ and $C_{2}$ are drawn. The tangent PQ touches $C_{1}$ at P and $C_{2}$ at Q. The tangent ST touches $C_{1}$ at S and $C_{2}$ at T. Mid points of the line segments PQ and ST are joined to form a line which meets the $x$ -axis at a point B. If $A B = \sqrt{5}$ , then the value of $r^{2}$ is [2023]

Smart Achievers · Accepted Answer

(2)

$C_{1} : x^{2} + y^{2} = 1 . . . . (i)$

$Let C_{2} : {(x - 4)}^{2} + {(y - 1)}^{2} = r^{2} . . . (ii)$

$Radical axis:$ $8 x + 2 y - 17 = 1 - r^{2} [from (i) and (ii)]$

$\Rightarrow 8 x + 2 y = 18 - r^{2}$

$B (\frac{18 - r^{2}}{8}, 0) and A (4, 1)$

$Given A B = \sqrt{5}$ $\Rightarrow \sqrt{{(\frac{18 - r^{2}}{8} - 4)}^{2} + 1} = \sqrt{5}$ $\Rightarrow r^{2} = 2$

Solution

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Solution

Ans. (2) C1: x2+y2=1 ....(i) Let C2: (x-4)2+(y-1)2=r2 ...(ii) Radical axis: 8x+2y-17=1-r2 [from (i) and (ii)] ⇒8x+2y=18-r2 B(18-r28,0) and A(4,1) Given AB=5⇒(18-r28-4)2+1=5⇒r2=2

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