Question

A circle S passes through the point (0, 1) and is orthogonal to the circles \[{{\left( x-1 \right)}^{2}}+{{y}^{2}}=16\] and \[{{x}^{2}}+{{y}^{2}}=1\], then
(This question has multiple correct options)
(a) The radius of S is 8
(b) The radius of S is 7
(c) Center of S is (– 7, 1)
(d) Center of S is (– 8, 1)

Answer 1

Hint: To solve this question, we should know the basic concept of the orthogonality condition which states that if two circles cut each other orthogonally having the equation of the circle as \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] and \[{{x}^{2}}+{{y}^{2}}+2{{g}^{'}}x+2{{f}^{'}}y+{{c}^{'}}=0\] follow the relation of \[2g\times{{g}^{'}}+2f\times{{f}^{'}}=c+{{c}^{'}}\].

Complete step-by-step solution -
We are given with two equations of the circle. Let us consider \[{{C}_{1}}\] be \[{{\left( x-1 \right)}^{2}}+{{y}^{2}}=16\] and \[{{C}_{2}}\] be \[{{x}^{2}}+{{y}^{2}}=1\]. We know that the circle which is cutting \[{{C}_{1}}\] and \[{{C}_{2}}\] orthogonally is S. So, let us consider the equation of S as \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]. As, it is given that the circle S satisfies (0, 1) point. So, we can write,
\[{{\left( 0 \right)}^{2}}+{{\left( 1 \right)}^{2}}+2g\left( 0 \right)+2f\left( 1 \right)+c=0\]
\[1+2f+c=0\]
\[c=-2f-1....\left( i \right)\]
We know that when two circles cut each other orthogonally having equation \[{{x}^{2}}+{{y}^{2}}+2{{g}^{'}}x+2{{f}^{'}}y+{{c}^{'}}=0\] and \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\] follow the relation \[2g\times{{g}^{'}}+2f\times{{f}^{'}}=c+{{c}^{'}}\].
Now, we have been given that
\[{{C}_{1}}:{{\left( x-1 \right)}^{2}}+{{y}^{2}}=16\]
We can write it as,
\[{{C}_{1}}:{{x}^{2}}+{{y}^{2}}-2x+1=16\]
\[{{C}_{1}}:{{x}^{2}}+{{y}^{2}}-2x-15=0\]
So, we can say \[{{g}_{1}}=-1\] and \[{{f}_{1}}=0\] and \[{{o}_{1}}=-15\].
As it is given that \[{{C}_{1}}\] and S cuts each other orthogonally, so we can write
\[2g\times {{g}_{1}}+2f\times {{f}_{1}}=C+{{C}_{1}}\]
Now, we will put the value of g, \[{{f}_{1}}\], \[{{C}_{1}}\] and C, we will get,
\[2g\left( -1 \right)+2f\left( 0 \right)=-2f-1-15\]
\[-2g+2f=-16\]
\[2g-2f=16\]
\[g-f=8....\left( ii \right)\]
We also have been given that
\[{{C}_{2}}:{{x}^{2}}+{{y}^{2}}=1\]
From this, we can say \[{{g}_{2}}=0\], \[{{f}_{2}}=0\] and \[{{C}_{2}}=-1\]. And \[{{C}_{2}}\] and S cuts each other orthogonally. So, we can write,
\[2g\times {{g}_{2}}+2f\times {{f}_{2}}=C+{{C}_{2}}\]
Now, we will put the value of \[{{g}_{2}}\], \[{{f}_{2}}\], \[{{C}_{2}}\] and C. So, we will get,
\[2g\left( 0 \right)+2f\left( 0 \right)=\left( -2f-1 \right)+\left( -1 \right)\]
\[\Rightarrow 0+0=-2f-1-1\]
\[2f+2=0\]
\[f=\dfrac{-2}{2}\]
\[f=-1....\left( iii \right)\]
Now, we will use the value of f from the equation (iii), we will get
\[g-\left( -1 \right)=8\] and \[C=-2\left( -1 \right)-1\]
\[g+1=8\] and \[C=2-1\]
g = 7 and C = 1
Hence, we will get the equation of circle S as
\[{{x}^{2}}+{{y}^{2}}+2\left( 7 \right)x+2\left( -1 \right)y+1=0\]
Now, we know that the radius of the circle \[=\sqrt{\left( {{g}^{2}}+{{f}^{2}}-C \right)}\] and center of the circle \[=\left( -g,-f \right)\]. So, we get a radius of S \[=\sqrt{{{7}^{2}}+{{\left( -1 \right)}^{2}}-1}\] and center of S is (– 7, 1).
The radius of S \[=\sqrt{49}\] and center of the circle is (– 7, 1)
The radius of S = 7 and center of S is (– 7, 1)
Hence, option (b) and (c) are the correct answers.

Note: We can also use the tangent method to find the radius and center of S but after a few steps that will also come to the same point but will complicate the solution. The possible mistake which we can make is by taking the wrong relation of orthogonality criteria or by taking the wrong values of f and g.