SATHEE: Circle 2 Question 16

Circle 2 Question 16

17. A circle $S$ passes through the point $(0, 1)$ and is orthogonal to the circles $(x - 1)^{2} + y^{2} = 16$ and $x^{2} + y^{2} = 1$ . Then,

(2014 Adv.)

(a) radius of $S$ is 8

(b) radius of $S$ is 7

(d) centre of $S$ is $(- 8, 1)$

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Let $A B C D$ be a square of side length 2 unit. $C_{2}$ is the circle through vertices $A, B, C, D$ and $C_{1}$ is the circle touching all the sides of square $A B C D . L$ is the line through $A$ .

(2006, 5M)

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Answer:

Correct Answer: (b, c)

Solution:

PLAN

(i) The general equation of a circle is

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

where, centre and radius are given by $(- g, - f)$ and $\sqrt{g^{2} + f^{2} - c}$ , respectively.

(ii) If the two circles $x^{2} + y^{2} + 2 g_{1} x + 2 f_{1} y + c_{1} = 0$ and $x^{2} + y^{2} + 2 g_{2} x + 2 f_{2} y + c_{2} = 0$ are orthogonal, then $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ .

Let circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

It passes through $(0, 1)$ .

$∴ 1 + 2 f + c = 0$

Orthogonal with $x^{2} + y^{2} - 2 x - 15 = 0$

$\begin{aligned} 2 g (- 1) & = c - 15 \\ \Rightarrow & = 15 - 2 g \\ Orthogonal with x^{2} + y^{2} - 1 & = 0 \\ c & = 1 \\ \Rightarrow g = 7 and f & = - 1 \end{aligned}$

Centre is $(- g, - f) \equiv (- 7, 1)$

$\begin{aligned} ∴ Radius & = \sqrt{g^{2} + f^{2} - c} \\ = \sqrt{49 + 1 - 1} = 7 \end{aligned}$

Circle 2 Question 16

17. A circle $S$ passes through the point $(0, 1)$ and is orthogonal to the circles $(x - 1)^{2} + y^{2} = 16$ and $x^{2} + y^{2} = 1$ . Then,

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Circle 2 Question 16

17. A circle S passes through the point (0,1) and is orthogonal to the circles (x−1)2+y2=16 and x2+y2=1. Then,

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17. A circle $S$ passes through the point $(0, 1)$ and is orthogonal to the circles $(x - 1)^{2} + y^{2} = 16$ and $x^{2} + y^{2} = 1$ . Then,