The equations of the tangents to the curve x^ | vedclass

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(B) Step $1$: The given curve is $x^2 + y^2 = 10$, which is a circle with center $(0, 0)$ and radius $r = \sqrt{10}$.Step $2$: The given line is $2x +

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$2x + y = \pm5\sqrt{2}$

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(B) Step $1$: The given curve is $x^2 + y^2 = 10$, which is a circle with center $(0, 0)$ and radius $r = \sqrt{10}$.Step $2$: The given line is $2x + y - 1 = 0$. The slope of this line is $m = -2$.Step $3$: Any line parallel to $2x + y - 1 = 0$ will have the form $2x + y + k = 0$.Step $4$: For a line $Ax + By + C = 0$ to be tangent to a circle $x^2 + y^2 = r^2$, the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $r$.Step $5$: The distance $d = \frac{|A(0) + B(0) + k|}{\sqrt{A^2 + B^2}} = \frac{|k|}{\sqrt{2^2 + 1^2}} = \frac{|k|}{\sqrt{5}}$.Step $6$: Setting $d = r$, we get $\frac{|k|}{\sqrt{5}} = \sqrt{10}$, which implies $|k| = \sqrt{50} = 5\sqrt{2}$.Step $7$: Thus, $k = \pm 5\sqrt{2}$. The equations of the tangents are $2x + y \pm 5\sqrt{2} = 0$, or $2x + y = \pm 5\sqrt{2}$.

The equations of the tangents to the curve $x^2 + y^2 = 10$, which are parallel to the line $2x + y - 1 = 0$, are:

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