Circle And System Of Circles Question 142
Question: The equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line $ y-4x+3=0 $ , is
[RPET 1985; MP PET 1989]
Options:
A) $ x^{2}+y^{2}+4x-10y+25=0 $
B) $ x^{2}+y^{2}-4x-10y+25=0 $
C) $ x^{2}+y^{2}-4x-10y+16=0 $
D) $ x^{2}+y^{2}-14y+8=0 $
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Answer:
Correct Answer: B
Solution:
First find the centre. Let centre be (h, k), then $ \sqrt{{{(h-2)}^{2}}+{{(k-3)}^{2}}}=\sqrt{{{(h-4)}^{2}}+{{(k-5)}^{2}}} $ ………….(i) and $ k-4h+3=0 $ ………….(ii)
From (i), we get
$ -4h-6k+8h+10k=16+25-4-9 $ or $ 4h+4k-28=0 $ or $ h+k-7=0 $ ………….(iii)
From (iii) and (ii), we get (h, k) as (2, 5).
Hence centre is (2, 5) and radius is 2. Now find the equation of circle.
Trick : Obviously, circle $ x^{2}+y^{2}-4x-10y+25=0 $ passes through (2, 3) and (4, 5).
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