If one end of a diameter of the circle x^2 + | vedclass

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Question

(A) The given equation of the circle is $x^2 + y^2 + 2x + 4y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g

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$(-3, -4)$

Vedclass · Answer

(A) The given equation of the circle is $x^2 + y^2 + 2x + 4y - 3 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 2 \implies g = 1$ and $2f = 4 \implies f = 2$. The center of the circle is $(-g, -f) = (-1, -2)$. Let the other end of the diameter be $(x_1, y_1)$. Since the center is the midpoint of the diameter,we have: $\frac{x_1 + 1}{2} = -1 \implies x_1 + 1 = -2 \implies x_1 = -3$. $\frac{y_1 + 0}{2} = -2 \implies y_1 = -4$. Thus,the other end of the diameter is $(-3, -4)$.

If one end of a diameter of the circle $x^2 + y^2 + 2x + 4y - 3 = 0$ is $(1, 0)$,then the other end of the diameter is:

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