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hkdsesecret

The circle should be x^2 + y^2 - 6 x + 6 sqrt(2) y + 24 = 0 .

Prog 4 :
The eqn.s of the tangents is y = - sqrt(2) x + 3 and y = - x / sqrt(2) .

Prog 1 :
The points of intersecton is ( 3 + sqrt(2) , 1 - 3sqrt(2) ) and ( 4 , - 2 sqrt(2) ) .

hkdsesecret

Sorry that only answer is provided for you because the steps is quite complicated. Here is an in¡Vc method.

Find the eqn.s of tangents to a circle froman external point
1.Let the eqn.s of tangents be y ¡V y_1 = m (x ¡V x_1 )
2.Simultaneous the equations
3.Putting discriminant=0

------------------------------------------------------------------------------

Let the eqn.s of tangents be y ¡V (¡V3) = m (x ¡V 3 sqrt(2) )
i.e. y = m x ¡V 3 sqrt(2) m ¡V 3

x^2 + y^2 ¡V 6 x + 6 sqrt(2) y + 24 = 0
x^2 + [ m x ¡V 3 sqrt(2) m ¡V 3 ]^2 ¡V 6 x + 6sqrt(2) [ m x ¡V 3 sqrt(2) m ¡V 3 ] + 24 = 0
x^2 + m^2 x^2¡V6 sqrt(2) m^2 x+18 m^2+6sqrt(2) m x¡V6 m x+18 sqrt(2) m¡V36 m¡V6 x¡V18 sqrt(2)+33 = 0
[ 1 +m^2 ] x^2 + [¡V6 sqrt(2) m^2+6 sqrt(2) m¡V6 m¡V6] x + [18 m^2+18 sqrt(2) m¡V36 m¡V18sqrt(2)+33] = 0

Discriminant = 0
[¡V6 sqrt(2) m^2+6 sqrt(2) m¡V6 m¡V6]^2 ¡V 4 [1+ m^2] [18 m^2+18 sqrt(2) m¡V36 m¡V18 sqrt(2)+33] = 0
72 sqrt(2) m^2¡V96 m^2¡V144 sqrt(2) m+216 m+72sqrt(2)¡V96 = 0
[72 sqrt(2)¡V96] m^2 + [¡V144 sqrt(2)+216] m +72sqrt(2)¡V96= 0
m= ¡V sqrt(2) or ¡V 1 / sqrt(2)

The eqn.s of the tangents is y = ¡V sqrt(2)x + 3 and y = ¡V x / sqrt(2) .

[ ¥»©«³Ì«á¥Ñ hkdsesecret ©ó 2016-8-12 04:12 PM ½s¿è ]