Statement 1: The only circle having radius sq | vedclass

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(A) For the circle $x^2 + y^2 - 6x + 2y = 0$,the center is $(3, -1)$ and the radius is $r = \sqrt{3^2 + (-1)^2 - 0} = \sqrt{10}$.Check if the center $

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Statement $1$ is false; Statement $2$ is true.

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(A) For the circle $x^2 + y^2 - 6x + 2y = 0$,the center is $(3, -1)$ and the radius is $r = \sqrt{3^2 + (-1)^2 - 0} = \sqrt{10}$.Check if the center $(3, -1)$ lies on the line $2x + y = 5$: $2(3) + (-1) = 6 - 1 = 5$. Since the center lies on the line,the line is a diameter (and thus a normal) to the circle. So,Statement $2$ is true.For Statement $1$,there are infinitely many circles with radius $\sqrt{10}$ whose centers lie on the line $2x + y = 5$. Any circle with center $(h, k)$ such that $2h + k = 5$ and radius $\sqrt{10}$ satisfies the condition. Thus,Statement $1$ is false.

Statement $1$: The only circle having radius $\sqrt{10}$ and a diameter along the line $2x + y = 5$ is $x^2 + y^2 - 6x + 2y = 0$.
Statement $2$: $2x + y = 5$ is a normal to the circle $x^2 + y^2 - 6x + 2y = 0$.

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Statement $1$: The only circle having radius $\sqrt{10}$ and a diameter along the line $2x + y = 5$ is $x^2 + y^2 - 6x + 2y = 0$.Statement $2$: $2x + y = 5$ is a normal to the circle $x^2 + y^2 - 6x + 2y = 0$.