Question

If two distinct chord, drawn from the point (p, q) on the circle \[{{x}^{2}}+{{y}^{2}}=px+qy\], where \[pq\ne 0\] are bisected by the x – axis, then
(a) \[{{p}^{2}}={{q}^{2}}\]
(b) \[{{p}^{2}}=8{{q}^{2}}\]
(c) \[{{p}^{2}}<8{{q}^{2}}\]
(d) \[{{p}^{2}}>8{{q}^{2}}\]

Answer 1

Hint: Draw a chord AB with coordinate A (p, q) which is bisected by the x-axis. Now, find the y – coordinate of B by using the midpoint formula given by: - \[y=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}\], where y is the midpoint of A and B, \[{{y}_{1}}\] and \[{{y}_{2}}\] are the y – coordinates of A and B respectively. Substitute the value of y – coordinate of B in the equation of a circle and form a quadratic equation in x. Substitute the value of this quadratic equation’s discriminant greater than 0 to get the required condition.

Complete step-by-step solution

In the above figure, we have assumed a chord AB with coordinates of A as (p, q) from which the chord is drawn. Here, M is the midpoint of AB.
Now, it is given that the chord is bisected on the x-axis. Therefore, M lies on the x-axis, and hence its y – coordinate must be O.
Now, let us find the y – coordinate of B. Let us assume it to be y. Then, by applying mid – point formula for the y – coordinate, we have,
y – coordinate of M = $\dfrac{\text{(y – coordinate of A + y – coordinate of B)}}{2}$
\[\begin{align}
  & \Rightarrow 0=\dfrac{q+y}{2} \\
 & \Rightarrow q=-y \\
 & \Rightarrow y=-q \\
\end{align}\]
Therefore, y – coordinate of B is –q. Since, this point lies on the circle. So, it will satisfy the equation of the circle. Substituting the value \[y=-q\] in the equation of circle, we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{\left( -q \right)}^{2}}=px+q.\left( -q \right) \\
 & \Rightarrow {{x}^{2}}+{{q}^{2}}=px-{{q}^{2}} \\
 & \Rightarrow {{x}^{2}}+2{{q}^{2}}-px=0 \\
 & \Rightarrow {{x}^{2}}-px+2{{q}^{2}}=0 \\
\end{align}\]
Now, in the question we have been given that there are two such chords which are bisected at x – axis. So, the above quadratic equation must provide two values of x. Therefore, the discriminant of the above quadratic equation must be greater than 0. So, mathematically,
\[{{b}^{2}}-4ac>0\], where b = co – efficient of x, a = coefficient of \[{{x}^{2}}\], c = constant term.
\[\begin{align}
  & \Rightarrow {{\left( -p \right)}^{2}}-4\times \left( 2{{q}^{2}} \right)\times 1>0 \\
 & \Rightarrow {{p}^{2}}-8{{q}^{2}}>0 \\
 & \Rightarrow {{p}^{2}}>8{{q}^{2}} \\
\end{align}\]
Hence, option (d) is the correct answer.

Note: One may note that in the question we have been given that there exist two such lines which are bisected by the x-axis and that is why we have substituted the value of discriminant of the quadratic equation greater than 0. If there would have been only one such chord then we would have substituted the value of discriminant equal to 0.