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)
Show Answer
Answer:
Correct Answer: 17. $16: 1$
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 $(0,1)$.
$$ \therefore \quad 1+2 f+c=0 $$
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 $(-g,-f) \equiv(-7,1)$
$$ \begin{aligned} \therefore \quad \text { Radius } & =\sqrt{g^{2}+f^{2}-c} \\ & =\sqrt{49+1-1}=7 \end{aligned} $$
