Question

From any point on the circle

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

tangents are drawn to the circle

x^{2} + y^{2} + 2 g x + 2 f y + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α = 0;

Prove that the angle between them is

2 α

.

Accepted Answer

The given circle is:
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0 \dots \dots . (1)$
Say $P \equiv (h, k)$ is any point on $(1)$ , then
$h^{2} + k^{2} + 2 g h + 2 f k + c = 0 \dots \dots \dots \dots (2)$
The other circle is:
$x^{2} + y^{2} + 2 g x + 2 f y + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α = 0$ $\dots \dots . (3)$
The radius of $(3)$ is:
$C A = \sqrt{(g^{2} + f^{2} - \{c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α\})}$
= $\sqrt{[g^{2} (1 - \cos^{2} α) + f^{2} (1 - \cos^{2} α) - c \sin^{2} α]}$
$= \sqrt{(g^{2} \sin^{2} α + f^{2} \sin^{2} α - c \sin^{2} α)}$
$= \sin α \sqrt{(g^{2} + f^{2} - c)}$
Again,

$P A =$ Length of the tangent drawn form point $P$ to the small circle given by $(3)$
$= \sqrt{\{h^{2} + k^{2} + 2 g h + 2 f k + c \sin^{2} α + (g^{2} + f^{2}) \cos^{2} α\}}$
$= \sqrt{\{h^{2} + k^{2} + 2 g h + 2 f k + c) - c \cos^{2} α + (g^{2} + f^{2}) \cos^{2} α)\}} \{∵ \sin^{2} α = (1 - \cos^{2} α)\}$
$= \sqrt{\{(g^{2} + f^{2} - c) \cos^{2} α\}} [∵ h^{2} + k^{2} + 2 g h + 2 f k + c = 0]$ by $(2)$
$= \cos α \sqrt{(g^{2} + f^{2} - c)}$
Now, $∠ A P B = 2 ∠ A P C$
and $\tan ∠ A P C = \frac{C A}{A P} = \frac{\sin α \sqrt{(g^{2} + f^{2} - c)}}{\cos α \sqrt{(g^{2} + f^{2} - a)}} = \tan α$
$∴ ∠ A P C = α$
So, $∠ A P B = 2 ∠ A P C = 2 α$

S L Loney Solutions for Chapter: The Circle, Exercise 4: EXAMPLES XXII

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: The Circle, Exercise 4: EXAMPLES XXII

Questions from S L Loney Solutions for Chapter: The Circle, Exercise 4: EXAMPLES XXII with Hints & Solutions

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