Question

The circle passing through \[\left( { - 1,0} \right)\] and touching the y -axis at \[\left( {0,2} \right)\] also passes through the point.
A) \[\left( { - \dfrac{3}{2},0} \right)\]
B) \[\left( { - \dfrac{5}{2},0} \right)\]
C) \[\left( { - \dfrac{3}{2},\dfrac{5}{2}} \right)\]
D) \[\left( { - 4,0} \right)\]

Answer 1

Hint: Standard form: If \[P\left( {x,{\text{ }}y} \right)\] is any point on the circle then the equation of a circle with centre at the origin and radius ‘r’ is given by
\[{x^2}{\text{ }} + {\text{ }}{y^2}{\text{ }} = {\text{ }}{r^2}\].
The above equation is the standard form of a circle.
Centre − radius form: If \[P\left( {x,{\text{ }}y} \right)\] is any point on the circle, then the equation of a circle with its centre at \[\left( {h,k} \right)\] and radius ‘r’ is given by distance formula
\[{(x - h)^2}{\text{ }} + {\text{ }}{(y - k)^2}{\text{ }} = {\text{ }}{r^2}\]
The above equation is the centre−radius form of equation of a circle
When a circle with centre \[\left( {h,k} \right)\] touches the x-axis, its radius =$\left| {y - coordinate\;of\;centre} \right| = \left| k \right|$
When a circle with centre \[\left( {h,k} \right)\] touches the y-axis, its radius =$\left| {x - coordinate\;of\;centre} \right| = \left| h \right|$

Complete step-by-step answer:
As the circle is touching the y-axis at \[\left( {0,2} \right)\]
Centre of the circle lies on the line y=2
Let the coordinates of the centre by (h,2), Radius of the circle is $\left| h \right|$
\[ \Rightarrow - h = \sqrt {{{\left( {h + 1} \right)}^2} + {{\left( {2 - 0} \right)}^2}} \]
as it passes through
\[
   \Rightarrow - h = \sqrt {{h^2} + 2h + 1 + 4} \\
   \Rightarrow {h^2} = {h^2} + 2h + 5 \\
   \Rightarrow h = - \dfrac{5}{2} \\
 \]
So the equation of the circle is
\[{\left( {x + \dfrac{5}{2}} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\dfrac{5}{2}} \right)^2}\]
By substitution the value of option A \[\left( { - \dfrac{3}{2},0} \right)\]in the above equation: -
\[
  {\left( {\left( { - \dfrac{3}{2}} \right) + \dfrac{5}{2}} \right)^2} + {\left( {\left( 0 \right) - 2} \right)^2} \\
   = {\left( {\dfrac{{ - 3 + 5}}{2}} \right)^2} + {\left( { - 2} \right)^2} \\
   = {\left( {\dfrac{2}{2}} \right)^2} + {2^2} \\
   = 1 + 4 = 5 \\
 \]
Which is not equal to RHS.
By substitution the value of option B \[\left( { - \dfrac{5}{2},0} \right)\]in the above equation: -
\[
  {\left( {\left( { - \dfrac{5}{2}} \right) + \dfrac{5}{2}} \right)^2} + {\left( {\left( 0 \right) - 2} \right)^2} \\
   = {\left( { - \dfrac{0}{2}} \right)^2} + {2^2} \\
   = 4 \\
 \]
Which is not equal to RHS.
By substitution the value of option C \[\left( { - \dfrac{3}{2},\dfrac{5}{2}} \right)\]in the above equation: -
\[
  {\left( {\left( { - \dfrac{3}{2}} \right) + \dfrac{5}{2}} \right)^2} + {\left( {\left( {\dfrac{5}{2}} \right) - 2} \right)^2} \\
   = {\left( {\dfrac{2}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^2} \\
   = 1 + \dfrac{1}{4} = \dfrac{5}{4} \\
 \]
Which is not equal to RHS.
By substitution the value of option D \[\left( { - 4,0} \right)\]in the above equation: -
\[
  {\left( {\left( { - 4} \right) + \dfrac{5}{2}} \right)^2} + {\left( {\left( 0 \right) - 2} \right)^2} \\
   = {\left( { - \dfrac{3}{2}} \right)^2} + {2^2} \\
   = \dfrac{9}{4} + 4 = \dfrac{{25}}{4} \\
 \]
Which is equal to RHS. \[\left( { - 4,0} \right)\] satisfies this equation.
which passes through \[\left( { - 4,0} \right)\]

So, option (D) is the correct answer.

Note: Sometimes it is easier to substitute the value of options given in the equation to check which one satisfies it then finding the solution. Even if you find the solution you can still substitute the same in the given equation to confirm the answer.