Question

Prove that from the point (a,b) on the circle ${x^2} + {y^2} - ax - by = 0$ two chords, each bisected by the axis of x can be drawn if ${a^2} > 8{b^2}$.

Answer 1

Here, we will proceed by using the Midpoint Theorem to find the coordinates of the midpoint of the chord to the given circle. Then, we will be putting the y-coordinate of the midpoint equal to zero because the chord is bisected by x-axis.

Complete step-by-step answer:
To prove- ${a^2} > 8{b^2}$
Consider a circle and there is a point A (a,b) on the circle as shown in the figure. Let the chord AB is drawn from the point A in such a way that both of this chord is bisected by x-axis.
Let us assume the coordinates of the point B be (c,d)
Given, equation of circle is ${x^2} + {y^2} - ax - by = 0{\text{ }} \to {\text{(1)}}$
Since, the chord AB is bisected by the x-axis which means that the midpoint of the chord AB should lie on x-axis. Let the midpoint of chord AB is M.
According to Midpoint Theorem, the midpoint of any line (AB) joining the points A (a,b) and B (c,d) is given by
x-coordinate of the midpoint = $\dfrac{{a + c}}{2}$ and y-coordinate of the midpoint = $\dfrac{{b + d}}{2}$
Using the above formulas, we can write
The coordinates of point M which is the midpoint of line AB are $\left( {\dfrac{{a + c}}{2},\dfrac{{b + d}}{2}} \right)$
For this midpoint M to lie on x-axis, the y-coordinate of the point M should be equal to zero
i.e., $\dfrac{{b + d}}{2} = 0 \\\ \Rightarrow b + d = 0 \\\ \Rightarrow d = - b \\\$
Since, point B (c,d) lies on the given circle so it will satisfy the equation of the given circle
By substituting x = c and y = d in equation (1), we get
$\Rightarrow {c^2} + {d^2} - ac - bd = 0$
By substituting d = -b in the above equation, we get

$$\Rightarrow {c^2} + {\left( { - b} \right)^2} - ac - b\left( { - b} \right) = 0 \\\ \Rightarrow {c^2} + {b^2} - ac + {b^2} = 0 \\\ \Rightarrow {c^2} - ac + 2{b^2} = 0{\text{ }} \to {\text{(2)}} \\\$$

The equation (2) represents a quadratic equation in variable c (where c is the x-coordinate of the end-point of a chord to the circle whose other end-point is A).
As we know that for any general quadratic equation in variable c i.e., $e{c^2} + fc + g = 0{\text{ }} \to {\text{(3)}}$, the solution is given by
$c = \dfrac{{ - f \pm \sqrt {{f^2} - 4eg} }}{{2e}}{\text{ }} \to {\text{(4)}}$
By comparing equations (2) and (3), we get
e = 1, f = -a and $g = 2{b^2}$
By putting the above values in equation (4), we get
$\Rightarrow c = \dfrac{{ - f \pm \sqrt {{{\left( { - a} \right)}^2} - 4\left( 1 \right)\left( {2{b^2}} \right)} }}{{2\left( 1 \right)}} \\\ \Rightarrow c = \dfrac{{ - f \pm \sqrt {{a^2} - 8{b^2}} }}{2} \\\$
For the quadratic equation (2) to have two real and positive roots, we discriminant should be positive (i.e., greater than 0)
i.e., ${a^2} - 8{b^2} > 0 \\\ \Rightarrow {a^2} > 8{b^2} \\\$
Clearly, the above inequality is the same inequality which needed to be proved.

Note: For any general quadratic equation $e{c^2} + fc + g = 0$, the discriminant is given by ${f^2} - 4eg$. For this quadratic equation to have two real and distinct roots, ${f^2} - 4eg > 0$, to have real and equal roots ${f^2} - 4eg = 0$ and to have imaginary roots ${f^2} - 4eg < 0$. Here, c represents the x-coordinate of the endpoint of the chord which needs to be real and distinct in order to have two required chords.