Question

A tangent is drawn to the circle

{(x - a)}^{2} + y^{2} = b^{2}

and a perpendicular tangent to the circle

{(x + a)}^{2} + y^{2} = c^{2}

find the locus of their point of intersection, and prove that the bisector of the angle between them always touches one or other of two fixed circle.

Accepted Answer

The equation of the circle is given as:
${(x - a)}^{2} + y^{2} = b^{2}$ $\dots (1)$
and ${(x + a)}^{2} + y^{2} = c^{2} \dots (2)$
Let $y = m x + k$ be any tangent to $(1)$ hence the perpendicular distance from the centre of $(1)$ must be equal to its radius.

So,

$\frac{|a m + k|}{\sqrt{1 + m^{2}}} = b$ [As center is $(a, 0)$ and radius is $b$ ]
or $k = - a m \pm b \sqrt{(1 + m^{2})}$
$∴ y = m (x - a) \pm b \sqrt{(1 + m^{2})} \dots . (3)$

Similarly, the equation of any tangent to $(2)$ will be

$∴ y = m^{'} (x + a) \pm b \sqrt{(1 + {(m^{'})}^{2})} \dots . (4)$
As by hypothesis, the two tangents are at right angles, $m m^{'} = - 1 \dots (5)$
The required locus will be obtained by eliminating $m$ and $m^{'}$ from equations $(3), (4)$ and $(5) .$
$y = - \frac{1}{m} (x + a) + c \sqrt{(1 + \frac{1}{m^{2}})}$
$o r m y = - (x + a) + c \sqrt{(1 + m^{2})}$ $\dots . . (6)$
Equation $(3)$ can be written as $b \sqrt{(1 + m^{2})} + m (x - a) - y = 0$
Equation $(6)$ can be written as $c \sqrt{(1 + m^{2})} - m y - (x + a) = 0$
Applying cross multiplication, we get:
$\frac{\sqrt{(1 + m^{2})}}{- (x^{2} - a^{2}) - y^{2}} = \frac{m}{- y c + b (x + a)} = \frac{1}{- b y - c (x - a)}$
$\frac{\sqrt{(1 + m^{2})}}{\sqrt{{(- y c + b x + b a)}^{2} + {(- b y - c x + a c)}^{2}}}$ [by ratio and proportion]
Squaring and simplifying, we get:
${(a^{2} - x^{2} - y^{2})}^{2} = {(b x - c y + a b)}^{2} + {(b y + c x - a c)}^{2}$ , which is the required locus.

Again equation of the bisector of $(3)$ and $(6)$ is
$\frac{m (x - a) - y + b \sqrt{(1 + m^{2})}}{\sqrt{(1 + m^{2})}} = \pm \frac{(x + a) + m y - c \sqrt{(1 + m^{2})}}{\sqrt{(1 + m^{2})}}$
Taking $+ v e$ sign, we get
$m x - m a - y + b \sqrt{(1 + m^{2})} = x + a + m y - c \sqrt{(1 + m^{2})}$
or $x (m - 1) - y (m + 1) - a (m + 1) + (b + c) \sqrt{(1 + m^{2})} = 0$
or $(y + a) (m + 1) = x (m - 1) + (b + c) \sqrt{(1 + m^{2})}$
or $y + a = (\frac{m - 1}{m + 1}) x + (\frac{b + c}{m + 1}) \sqrt{(1 + m^{2})}$

EXAMPLES XXII

S L Loney Mathematics Solutions for Exercise - EXAMPLES XXII

Questions from EXAMPLES XXII with Hints & Solutions

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