Question

Find the condition that the straight line \[cx-by+{{b}^{2}}=0\] may touch the circle \[{{x}^{2}}+{{y}^{2}}=ax+by\] and find the point of contact.

Answer 1

Hint: We will substitute the value of y from the line equation in the equation of circle because the line is the tangent to the circle and solving this we will get the condition when the line touches the circle.

Complete step-by-step answer:
The equation of the circle mentioned in the question is \[{{x}^{2}}+{{y}^{2}}=ax+by\] and it is given that the line \[cx-by+{{b}^{2}}=0\] touches the circle which means that the line here is a tangent.
Equation of the circle is \[{{x}^{2}}+{{y}^{2}}=ax+by........(1)\]
Equation of the line is \[cx-by+{{b}^{2}}=0......(2)\]
Rearranging equation (2) and solving for y we get,
\[\begin{align}
  & \Rightarrow by=cx+{{b}^{2}} \\
 & \Rightarrow y=\dfrac{cx+{{b}^{2}}}{b}......(3) \\
\end{align}\]
Now substituting the value of y from equation (3) in equation (1) because the line is the tangent to the circle and hence we get,
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{cx+{{b}^{2}}}{b} \right)}^{2}}-ax-b\left( \dfrac{cx+{{b}^{2}}}{b} \right)=0........(4)\]
Cancelling similar terms and rearranging equation (4) we get,
\[\Rightarrow {{x}^{2}}+{{\left( \dfrac{cx+{{b}^{2}}}{b} \right)}^{2}}-ax-cx-{{b}^{2}}=0........(5)\]
Now taking b as LCM and expanding the squares in equation (5) we get,
\[\Rightarrow {{b}^{2}}{{x}^{2}}+{{c}^{2}}{{x}^{2}}+{{b}^{4}}+2cx{{b}^{2}}-ax{{b}^{2}}-cx{{b}^{2}}-{{b}^{4}}=0........(6)\]
Now again cancelling similar terms and rearranging in terms of a quadratic function we get,
\[\begin{align}
  & \Rightarrow {{b}^{2}}{{x}^{2}}+{{c}^{2}}{{x}^{2}}+cx{{b}^{2}}-ax{{b}^{2}}=0 \\
 & \Rightarrow ({{b}^{2}}+{{c}^{2}}){{x}^{2}}+(c{{b}^{2}}-a{{b}^{2}})x=0......(7) \\
\end{align}\]
Now for the line to be the tangent, the above quadratic equation (7) must have only one solution. So using this information we get,
\[\begin{align}
  & \Rightarrow c{{b}^{2}}-a{{b}^{2}}=0 \\
 & \Rightarrow {{b}^{2}}(a-c)=0.......(8) \\
\end{align}\]
Since b is not equal to 0 in equation (8) we get \[a=c\]. So the condition that the straight line touches the circle is \[a=c\].
Now \[({{b}^{2}}+{{c}^{2}}){{x}^{2}}=0\] so x is equal to 0.
Substituting x equal to 0 in equation (2) we get,
\[\Rightarrow c(0)-by+{{b}^{2}}=0......(9)\]
Now solving for y in equation (9) we get,
\[\begin{align}
  & \Rightarrow by={{b}^{2}} \\
 & \Rightarrow y=b \\
\end{align}\]
So the point of contact between the circle and the tangent is (0, b) as x is equal to 0 and y is equal to b.

Note: Remembering the definition of the tangent is the key here. We in a hurry can commit a mistake in solving equation (5) and equation (7) and hence we need to be careful while doing this. Remember we are substituting the value of x in the line equation to get the value of y and hence we will get the coordinates of the point of contact.