The point diametrically opposite to the point P(1,0) on the circle x2+y2+2 x+4 y-3=0 is

Question

The point diametrically opposite to the point P(1,0) on the circle x2+y2+2 x+4 y-3=0 is (A) (3,-4) (B) (-3,4) (C) (-3,-4) (D) (3,4)

Tardigrade Admin · Accepted Answer

The given circle is x2+y2+2 x+4 y-3=0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/3b90592ce9af8f0c2c241d419785332f-.png" /> Centre (-1,-2) Let Q(α, β) be the point diametrically opposite to the point P(1,0) then (1+α/2)=-1 and (0+β/2)=-2 ⇒ α=-3, β=-4 So, Q is (-3,-4)

Q. The point diametrically opposite to the point $P (1, 0)$ on the circle $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$ is

$(3, - 4)$

$(- 3, 4)$

$(- 3, - 4)$

$(3, 4)$

The given circle is $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$

Centre $(- 1, - 2)$
Let $Q (α, β)$ be the point diametrically opposite to
the point $P (1, 0)$
then $\frac{1 + α}{2} = - 1$ and $\frac{0 + β}{2} = - 2$
$\Rightarrow α = - 3, β = - 4$
So, $Q$ is $(- 3, - 4)$

Q. The point diametrically opposite to the point P(1,0) on the circle x2+y2+2x+4y−3=0 is

(3,−4)

(−3,4)

(−3,−4)

(3,4)