Circle And System Of Circles Question 15
Question: The equation of the tangent to the circle $ x^{2}+y^{2}=a^{2} $ which makes a triangle of area $ a^{2} $ with the co-ordinate axes, is
Options:
A) $ x\pm y=a\sqrt{2} $
B) $ x\pm y=\pm a\sqrt{2} $
C) $ x\pm y=2a $
D) $ x+y=\pm 2a $
Show Answer
Answer:
Correct Answer: B
Solution:
Let the tangent be of form $ \frac{x}{x_1}+\frac{y}{y_1}=1 $ and area of $ \Delta $ formed by it with coordinate axes is $ \frac{1}{2}x_1y_1=a^{2} $ ………….(i)
Again, $ y_1x+x_1y-x_1y_1=0 $
Applying conditions of tangency $ | \frac{-x_1y_1}{\sqrt{x_1^{2}+y_1^{2}}} |\ =a $ or $ (x_1^{2}+y_1^{2})=\frac{x_1^{2}y_1^{2}}{a^{2}} $ ………….(ii)
From (i) and (ii), we get $ x_1,y_1 $ ; which gives equation of tangent as $ x\pm y=\pm a\sqrt{2} $ . Trick: There may be 4 tangents (as in figure). As the lines $ x\pm y=\pm a\sqrt{2} $ make triangle of area -a- in all four quadrants.
BETA
