What will be the equation of the tangent to the circle x^2 + y^2 – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?

Category: QuestionsWhat will be the equation of the tangent to the circle x^2 + y^2 – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?
Editor">Editor Staff asked 11 months ago

What will be the equation of the tangent to the circle x^2 + y^2 – 6x + 4y – 7 = 0, which are perpendicular to the straight line 2x – y + 3 = 0?
 
(a) x – 2y + 11 = 0
 
(b) x – 2y – 11 = 0
 
(c) x + 2y + 11 = 0
 
(d) x + 2y – 11 = 0
 
I got this question in a national level competition.
 
This intriguing question originated from Calculus Application topic in portion Application of Calculus of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

The correct choice is (c) x + 2y + 11 = 0
 
To explain I would say: The equation of any straight line perpendicular to the line 2x – y + 3 = 0 is,
 
x + 2y + k = 0   ……….(1)
 
Now, the co-ordinate of the center of the circle (3, -2) and its radius is,
 
√(9 + 4 – (-7) = 2√5
 
If straight line (1) be tangent to the given circle then, the perpendicular distance of the point (3, -2) from the line (1) = radius of the circle
 
Thus, ±(3 + 2(-2) + k)/√(1 + 4)
 
Or k – 1 = 2√5 * √5
 
So, k = 1 ± 10
 
= 11 or -9
 
Putting the value of k in (1) we get,
 
x + 2y + 11 = 0 and x + 2y – 9 = 0