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Miscellaneous Exercise 6 (II) · Q70

Q.Find the equations of the tangents to the circle x2+y2=36x^2+y^2=36 which are perpendicular to the line 5x+y=25x+y=2.

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Line 5x+y=25x+y=2 has slope −5-5; a line perpendicular to it has slope m=15m=\dfrac15. For x2+y2=36x^2+y^2=36 (r2=36r^2=36): c=±36(125)+36=±3625+36=±93625=±6265c=\pm\sqrt{36\left(\frac1{25}\right)+36}=\pm\sqrt{\frac{36}{25}+36}=\pm\sqrt{\frac{936}{25}}=\pm\frac{6\sqrt{26}}{5} (since 936=36×26936=36\times26, 936=626\sqrt{936}=6\sqrt{26}). So $y=\dfrac15x\pm\dfrac{6\sqrt{26} …

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