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Miscellaneous Exercise 6 (II) · Q71

Q.Find the equations of the tangents to the circle x2+y2−2x+8y−23=0x^2+y^2-2x+8y-23=0 having slope 33.

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Circle: x2+y2−2x+8y−23=0x^2+y^2-2x+8y-23=0, so g=−1,f=4,c=−23g=-1,f=4,c=-23; centre (1,−4)(1,-4), radius 1+16+23=40=210\sqrt{1+16+23}=\sqrt{40}=2\sqrt{10}. Let X=x−1,Y=y+4X=x-1,Y=y+4 (shifting the centre to the origin), so the circle becomes X2+Y2=40X^2+Y^2=40. A tangent with slope 33 in these shifted coordinates: c′=±40(9)+40=±400=±20c'=\pm\sqrt{40(9)+40}=\pm\sqrt{400}=\pm20, giving Y=3X±20Y=3X\pm20. Substituting …

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