The point diametrically opposite to the point P(1,0) on the circle x2+y2+2x+4y−3=0 is:
Solutions · 1
Official solution
The given circle is x2+y2+2x+4y−3=0
Center (−1,−2)
Let Q \[\left( {\alpha ,\beta } \right)\] be the point diametrically opposite to the point P(1,0),
then \[{{1 + \alpha } \over 2} = - 1\] and \[{{0 + \beta } \over 2} = - 2\]
\[ \Rightarrow \alpha = - 3,\beta = - 4,\] So, Q is \[\left( { - 3, - 4} \right)\]