Prove that the line $y = m x + c$ will be a tangent to the circle $(x - a)^{2} + (y - b)^{2} = r^{2}$ if $m^{2} (a^{2} - r^{2}) + 2 m a (c - b) + (c - b)^{2} = r^{2}$ .

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Solution

Perpendicular from the centre $(a, b)$ to the tangent $m x - y + c = 0$ should be equal to radius $r$
$\frac{m a - b + c}{\sqrt{(m^{2} + 1)}} = r$ . Square
$m^{2} a^{2} + (c - b)^{2} + 2 m a (c - b) = r^{2} (m^{2} + 1)$
or $m^{2} (a^{2} - r^{2}) + 2 m a (c - b) + (c - b)^{2} = r^{2}$ .

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r (< a)

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)

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

A

θ = α

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

(x - a) cos α + (y - b) sin α = r

(a,b) is the mid point of the chord $¯ A B$ of the circle $x^{2} + y^{2} = r^{2}$ . The tangent at A,B meet a C. then area of $Δ A B C$

x = a r sin θ cos ϕ

y = b r sin θ sin φ

z = c r cos θ

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = r^{2}

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