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Question

# Equation of the tangent to the circle, at the point (1,–1), whose centre is the point of intersection of the straight lines x–y=1 and 2x+y=3 is :

A
x+4y+3=0
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B
x3y4=0
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C
4x+y3=0
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D
3xy4=0
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Solution

## The correct option is A x+4y+3=0 Solving equation x−y=1 & 2x+y=3, we get, x=43,y=13 ⇒ Centre of the circle=(43,13) Slope of radius (OP)=−1−131−43=4 Since, tangent is perpendicular to radius. ∴ Slope of tangent=−14 ⇒ Equation of the tangent at (1,−1) is, y+1=−14(x−1)⇒x+4y+3=0 OR, Equation of the circle is, (x−43)2+(y−13)2=(1−43)2+(−1−13)2⇒3x2+3y2−8x−2y=0 Hence, equation of the tangent at (1,−1) is, 3x−3y−4(x+1)−(y−1)=0⇒x+4y+3=0

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