Circle And System Of Circles Question 118
Question: The equation of the circle passing through the point $ (-1,\ -3) $ and touching the line $ 4x+3y-12=0 $ at the point (3, 0), is
Options:
A) $ x^{2}+y^{2}-2x+3y-3=0 $
B) $ x^{2}+y^{2}+2x-3y-5=0 $
C) $ 2x^{2}+2y^{2}-2x+5y-8=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let the equation be $ x^{2}+y^{2}+2gx+2fy+c=0 $ ………….(i) But it passes through $ (-1,\ -3) $ and (3, 0) therefore $ 10-2g-6f+c=0 $ ………….(ii) $ 9+6g+c=0 $ ………….(iii) Also centre is $ C(-g,\ -f) $ . Slope of tangents $ =-\frac{4}{3}\Rightarrow $ Slope of normal $ =\frac{3}{4} $
$ \Rightarrow \frac{f}{3+g}=\frac{3}{4}\Rightarrow 3g-4f+9=0 $ …..(iv) Now on
Solving (ii), (iii) and (iv), we get $ g=-1,\ f=\frac{3}{2} $ and $ c=-3 $ Therefore, the equation of circle is $ x^{2}+y^{2}-2x+3y-3=0 $ . Trick : The points (-1, -3) and (3, 0) must satisfy the equation of circle. Circle given in (a) satisfies both the points. Also check whether it touches the line $ 4x+3y-12=0 $ or not.
BETA
