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Circle

Section: Mathematics

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Question : 20 of 27

Marks: +1, -0

The locus of the mid–point of the chord of contact of tangents drawn from points lying on the straight line 4x – 5y = 20 to the circle

x^2 + y^2

[JEE Adv 2012 P1]

$20(x^2 + y^2 ) - 36x + 45y = 0$
$20(x^2 + y^2 ) + 36x - 45y = 0$
$36(x^2 + y^2 ) - 20x + 45y = 0$
$36(x^2 + y^2 ) + 20x - 45y = 0$

Solution:

Equation of the chord bisected at P(h, k)

hx + ky =

h^2 + k^2

(1)

Let any point on line be

\left(\alpha, \frac{4}{5} (\alpha-4)\right)

.

Equation of the chord of contact is

\alpha x + \left(\frac{4}{5}(\alpha-4)\right) y

= 9 (2)

Comparing Eqs. (1) and (2), we get

\frac{h}{\alpha}

=

\frac{k}{\frac{4}{5}(\alpha-4)}

=

\frac{h^2+k^2}{9}

α =

\frac{20h}{4h-5k}

Now,

\frac{h(4h-5k)}{20h}

=

\frac{h^2+k^2}{9}

20 (h^2+k^2)

= 9 (5h - 5k)

20(x^2+ y^2) - 36x + 45y = 0

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