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Question

The tangent from the point of intersection of the lines $$2x-3y+1=0$$ and $$3x-2y-1=0$$ to the circle $${ x }^{ 2 }+{ y }^{ 2 }+2x-4y=0$$ is-


A
x+2y=0,x2y+1=0
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B
2xy1=0
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C
y=x,y=3x2
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D
2x+y+1=0
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Solution

The correct option is B $$2x-y-1=0$$
Given lines 
$$2x-3y+1=0$$     ....(1)
$$3x-2y-1=0$$     .....(2)

Solving (1) and (2), we get
$$x=1, y=1$$
So, the point of the intersection of the lines is $$(1,1)$$
Given equation of circle is $$x^2+y^2+2x-4y=0$$
Here, $$g=1,f=-2,c=0$$
So, eqn of tangent from $$(1,1)$$ to the circle is
$$x+y+1(x+1)-2(y+1)=0$$
$$\Rightarrow 2x-y-1=0$$

Maths

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