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Miscellaneous Exercise 6 (II) · Q66

Q.Show that 2x+y+6=02x+y+6=0 is a tangent to x2+y2+2x−2y−3=0x^2+y^2+2x-2y-3=0. Find its point of contact.

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Circle: x2+y2+2x−2y−3=0x^2+y^2+2x-2y-3=0, so g=1,f=−1,c=−3g=1,f=-1,c=-3; centre =(−1,1)=(-1,1), radius =1+1+3=5=\sqrt{1+1+3}=\sqrt5. Perpendicular distance from (−1,1)(-1,1) to the line 2x+y+6=02x+y+6=0: ∣2(−1)+1+6∣4+1=∣−2+1+6∣5=55=5=\dfrac{|2(-1)+1+6|}{\sqrt{4+1}}=\dfrac{|{-2+1+6}|}{\sqrt5}=\dfrac{5}{\sqrt5}=\sqrt5= radius. So the line is tangent. …

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