The equation of tangents to the circle x^2+y^ | vedclass

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(A) The equation of the circle is $x^2+y^2=4$,so the radius $r = 2$. The given line is $x+2y+3=0$,which can be written as $y = -\frac{1}{2}x - \frac{3

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$x+2y = \pm 2\sqrt{5}$

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(A) The equation of the circle is $x^2+y^2=4$,so the radius $r = 2$. The given line is $x+2y+3=0$,which can be written as $y = -\frac{1}{2}x - \frac{3}{2}$. The slope of this line is $m = -\frac{1}{2}$. Since the tangents are parallel to this line,their slope is also $m = -\frac{1}{2}$. The equation of a tangent to the circle $x^2+y^2=r^2$ with slope $m$ is given by $y = mx \pm r\sqrt{1+m^2}$. Substituting $m = -\frac{1}{2}$ and $r = 2$: $y = -\frac{1}{2}x \pm 2\sqrt{1 + (-\frac{1}{2})^2}$ $y = -\frac{1}{2}x \pm 2\sqrt{1 + \frac{1}{4}}$ $y = -\frac{1}{2}x \pm 2\sqrt{\frac{5}{4}}$ $y = -\frac{1}{2}x \pm 2 	imes \frac{\sqrt{5}}{2}$ $y = -\frac{1}{2}x \pm \sqrt{5}$ Multiplying by $2$: $2y = -x \pm 2\sqrt{5}$ $x+2y = \pm 2\sqrt{5}$

The equation of tangents to the circle $x^2+y^2=4$ which are parallel to $x+2y+3=0$ are

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