Question
If P(2,8) is an interior point of a circle x2+y2−2x+4y−p=0 which neither touches nor intersects the axes, then set for p is -
Answer: Option D
:
D
Since the point P is interior to the circle,S1 < 0
=(22)+(82)−2.(2)+4(8)−p<0
=96−p<0
=p>96
Also given that the circle doesn't touches any of the axes.
So, g2−c < 0
f2−c < 0
g2−c < 0
=1+p<0
=p<−1
Also,
f2−c < 0
=4+p<0
=p<−4
Since p<−4 and p>96 is not possible, the set for p will be a null set meaning it'll not have any element in it.
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:
D
Since the point P is interior to the circle,S1 < 0
=(22)+(82)−2.(2)+4(8)−p<0
=96−p<0
=p>96
Also given that the circle doesn't touches any of the axes.
So, g2−c < 0
f2−c < 0
g2−c < 0
=1+p<0
=p<−1
Also,
f2−c < 0
=4+p<0
=p<−4
Since p<−4 and p>96 is not possible, the set for p will be a null set meaning it'll not have any element in it.
Was this answer helpful ?
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