SATHEE: Circle 5 Question 20

Circle 5 Question 20

20. Let a circle be given by

$2 x (x - a) + y (2 y - b) = 0, (a \neq 0, b \neq 0)$

Find the condition on $a$ and $b$ if two chords, each bisected by the $X$ -axis can be drawn to the circle from $(a, b / 2)$ .

$(1992, 6 M)$

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Answer:

Correct Answer: 20. $a^{2} > 2 b^{2}$

Solution:

The given circle can be rewritten as

$x^{2} + y^{2} - a x - \frac{b y}{2} = 0$

Let one of the chord through $(a, b / 2)$ be bisected at $(h, 0)$ .

Then, the equation of the chord having $(h, 0)$ as mid-point is

$\begin{matrix} T = S_{1} \\ \Rightarrow h \cdot x + 0 \cdot y - \frac{a}{2} (x + h) - \frac{b}{4} (y + 0) = h^{2} + 0 - a h - 0 \\ \Rightarrow h - \frac{a}{2} x - \frac{b y}{4} - \frac{a}{2} h = h^{2} - a h \end{matrix}$

It passes through $(a, b / 2)$ , then

$\begin{aligned} h - \frac{a}{2} a - \frac{b}{4} \cdot \frac{b}{2} - \frac{a}{2} h = h^{2} - a h \\ \Rightarrow & h^{2} - \frac{3}{2} a h + \frac{a^{2}}{2} + \frac{b^{2}}{8} = 0 \end{aligned}$

According to the given condition, Eq. (iii) must have two distinct real roots. This is possible, if the discriminant of Eq. (iii) is greater than 0 .

i.e. $\frac{9}{4} a^{2} - 4 \frac{a^{2}}{2} + \frac{b^{2}}{8} > 0 \Rightarrow \frac{a^{2}}{4} - \frac{b^{2}}{2} > 0$

$\Rightarrow a^{2} > 2 b^{2}$

Circle 5 Question 20

20. Let a circle be given by

Answer:

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Circle 5 Question 20

20. Let a circle be given by

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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