Circle And System Of Circles Question 188
Question: The equation of the circle whose diameter lies on $ 2x+3y=3 $ and $ 16x-y=4 $ which passes through (4,6) is
[Kurukshetra CEE 1998]
Options:
A) $ 5(x^{2}+y^{2})-3x-8y=200 $
B) $ x^{2}+y^{2}-4x-8y=200 $
C) $ 5(x^{2}+y^{2})-4x=200 $
D) $ x^{2}+y^{2}=40 $
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Answer:
Correct Answer: A
Solution:
Let point $ (x_1,\ y_1) $ on the diameter.
$ \Rightarrow 2x_1+3y_1=3 $ ………….(i) $ 16x_1-y_1=4 $ ………….(ii) On
Solving (i) and (ii), we get centre,
$ \Rightarrow x_1=\frac{3}{10},\ y_1=\frac{4}{5} $
$ \therefore $ Equation of circle, $ {{(x-x_1)}^{2}}+{{(y-y_1)}^{2}}=r^{2}\Rightarrow {{( x-\frac{3}{10} )}^{2}}+{{( y-\frac{4}{5} )}^{2}}=r^{2} $
$ \because $ Circle passes through (4, 6).
So, $ r^{2}={{( \frac{37}{10} )}^{2}}+{{( \frac{26}{5} )}^{2}}\Rightarrow r^{2}=\frac{4073}{100} $
$ \therefore $ Required equation of circle is $ {{( x-\frac{3}{10} )}^{2}}+{{( y-\frac{4}{5} )}^{2}}=\frac{4073}{100} $
$ \Rightarrow 5(x^{2}+y^{2})-3x-8y=200 $ .
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