Gujarati
10-1.Circle and System of Circles
medium

The equation of the tangents to the circle ${x^2} + {y^2} + 4x - 4y + 4 = 0$ which make equal intercepts on the positive coordinate axes is given by

A

$x + y + 2\sqrt 2 = 0$

B

$x + y = 2\sqrt 2 $

C

$x + y = 2$

D

None of these

Solution

(b) Equation of tangent is of the form $x + y + c = 0$ and also it obeys condition of tangency,

$i.e.$, $\left| {\frac{{ – 2 + 2 + c}}{{\sqrt 2 }}} \right| = \sqrt {4 + 4 – 4} $

$\Rightarrow c = \pm \,2\sqrt 2 $

But for positive intercepts, $c = – 2\sqrt 2 $

 The tangent is $x + y = 2\sqrt 2 $.

Standard 11
Mathematics

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