Circle And System Of Circles Question 128
Question: A circle has radius 3 units and its centre lies on the line $ y=x-1 $ . Then the equation of this circle if it passes through point (7,3), is
[Roorkee 1988]
Options:
A) $ x^{2}+y^{2}-8x-6y+16=0 $
B) $ x^{2}+y^{2}+8x+6y+16=0 $
C) $ x^{2}+y^{2}-8x-6y-16=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let its centre be (h, k), then $ h-k=1 $ … (i) Also radius $ a=3 $ Equation is $ {{(x-h)}^{2}}+{{(y-k)}^{2}}=9 $ Also it passes through (7, 3)
i.e., $ {{(7-h)}^{2}}+{{(3-k)}^{2}}=9 $ ………….(ii)
We get h and k from (i) and (ii)
Solving simultaneously as (4, 3).
Equation is $ x^{2}+y^{2}-8x-6y+16=0 $ .
Trick : Since the circle $ x^{2}+y^{2}-8x-6y+16=0 $ satisfies the given conditions.
BETA
