# the point diametrically opposite to the point p(1,0) on the circle x2+y2+2x+3y

Anoopam Mishra
126 Points
7 years ago
Let the point be (h,k), then the centre of the circle is $((h+1)/2,k/2)$
But the centre of the circle is (-2,-3) from the givern equation.
=> $(h+1)/2 = -2$ => h=-5
and $k/2 = -3$ => k=-6.
Hence the diametrically opposite point is $(-5,-6)$
Raunaq Mehta
39 Points
6 years ago
Let the point be (h,k), then the centre of the circle is ((h+1)/2,k/2)But the centre of the circle is (-2,-3) from the givern equation.=> (h+1)/2 = -2 => h=-5and k/2 = -3 => k=-6.Hence the diametrically opposite point is (-5,-6)
Shubh mehta
19 Points
5 years ago
Let the point be (h,k), then the centre of the circle is ((h+1)/2,k/2)But the centre of the circle is (-2,-3) from the givern equation.=> (h+1)/2 = -2 => h=-5and k/2 = -3 => k=-6.so the diametrically opposite point is (-5,-6)
Abrar gani
19 Points
5 years ago
X2+y2 +2x+4y-3=oC(-1,-2)As P and Q are diametrically opposite therefore point C is the midpoint of line joining PandeyLetQ(x, y)x+1/2=1x=-3andy+0/2=-2Y=-4Implies that Point Q(-3,-4)