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
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.
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