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

(c) centre of $S$ is $(-7,1)$

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

Passage Based Problems

Passage

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:

  1. 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 $(0,1)$.

$ \therefore \quad 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 \\ \text { Orthogonal with } \quad x^{2}+y^{2}-1 & =0 \\ c & =1 \\ \Rightarrow \quad g=7 \text { and } f & =-1 \end{aligned} $

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

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



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