Question

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,

• A. Radius of S is 8
• B. Radius of S is 7
• C. Centre of S is (-7, 1)
• D. Center of S is (-8, 1)

Answer 1

Answer: (B), (C)

Centre of S is (-7, 1)

(a)The general equation of a circle is
$x^2+y^2+2gx+2fy+c=0$
Where, centre and radius are given by $(-g,-f)$ and $\sqrt{g^2+f^2-c}$, respectively.

(b) If the two circles $x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ and $x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$ are orthogonal, then $2g_1{g}_2+2f_1{f}_2=c_1+c_2$.

Let the circle be $x^2+y^2+2gx+2fy+c=0$
It passes through (0, 1).
$\therefore 1+2f+c=0$ ....(i)
Orthogonal with
$x^2+y^2-2x-15=0\therefore 2g(-1)=c-15usingproperty\left(b\right)statedabove$
$\Rightarrow c=15-2g$ ....(ii)
Similarly,
Orthogonal with $x^2+y^2-1=0$
$thereforec=1$ .....(iii)
Solving (i), (ii) and (iii), we get
$\Rightarrow g=7$ and $f=-1$
Centre is $(–g,-f)=(-7,1)$
$\therefore$ Radius = $\sqrt{g^2+f^2-c}$
$=\sqrt{49+1-1}=7$