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Exercise 6.3 · Q34

Q.Find the equation of the tangent to the circle x2+y2−4x+3y+2=0x^2+y^2-4x+3y+2=0 at the point (4,−2)(4,-2).

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The circle is x2+y2−4x+3y+2=0x^2+y^2-4x+3y+2=0, so g=−2,f=32,c=2g=-2,f=\dfrac32,c=2; check (4,−2)(4,-2) lies on it: 16+4−16−6+2=016+4-16-6+2=0. Confirmed. Applying the tangent formula at (x1,y1)=(4,−2)(x_1,y_1)=(4,-2):

x(4)+y(−2)+(−2)(x+4)+32(y−2)+2=0x(4)+y(-2)+(-2)(x+4)+\frac32(y-2)+2=0

4x−2y−2x−8+32y−3+2=04x-2y-2x-8+\frac32y-3+2=0 …

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