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 = 0I 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

What will be the equation of the tangent to the circle x^2 + y^2 – 6

1.	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 = 0I 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
Answer» 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

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