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Miscellaneous Exercise 6 (I) · Q38

Q.The equations of the tangents to the circle x2+y2=4x^2+y^2=4 which are parallel to x+2y+3=0x+2y+3=0 are (A) x−2y=2x-2y=2 (B) x+2y=±23x+2y=\pm2\sqrt3 (C) x+2y=±25x+2y=\pm2\sqrt5 (D) x−2y=±25x-2y=\pm2\sqrt5

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The line x+2y+3=0x+2y+3=0 has slope m=−12m=-\dfrac12. For the circle x2+y2=4x^2+y^2=4 (r2=4r^2=4), a tangent of this slope satisfies c=±r2m2+r2=±4⋅14+4=±1+4=±5c=\pm\sqrt{r^2m^2+r^2}=\pm\sqrt{4\cdot\frac14+4}=\pm\sqrt{1+4}=\pm\sqrt5. So the tangent lines are y=−12x±5y=-\dfrac12x\pm\sqrt5; multiplying by 22: 2y=−x±252y=-x\pm2\sqrt5, …

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