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The straight line2x +y=21 meets the coordinate axes at A and B. A circle is drawn through A B, and the origin. Then, the sum of perpendicular distances from A and B on the tangent to the circle at the origin is

 

Option: 1

1


Option: 2

2


Option: 3

\sqrt{(5)} / 2


Option: 4

0


Answers (1)

best_answer

When the line intersects the x-axis, y=0

2x+0=2

x=1

Therefore, point A has coordinates(1,0)

When the line intersects the y-axis(x=0)

\begin{aligned} & \quad 2(0)+y=2 \\ & y=2 \end{aligned}

Therefore, point B has coordinates (2,0)

The equation of the circle that passes through points A, B, and the origin (0,0) (\text { center } x, \text { centery })=((1+0) / 2,(0+2) / 2)=(1 / 2,1)

The radius of the circle is 


r=\sqrt{\left((1 / 2-1)^2+(1-0)^2\right)}=\sqrt{(1 / 4+1)}=\sqrt{(5)} / 2

Equation of the circle is:

\begin{gathered} \left.(x-1 / 2)^2+(y-1)^2=(\sqrt{(} 5) / 2\right)^2 \\ (x-1 / 2)^2+(y-1)^2=5 / 4 \end{gathered}

tangent line to the circle at the origin(0,0)

2(x-1 / 2)+2(y-1) y^{\prime}=0

At the point (0,0)his becomes:

2(-1 / 2) y^{\prime}=0

The slope of the tangent line at the origin is  y^{\prime}=0

 

Since the line is horizontal, its equation is simply y = 0.

 The perpendicular distance from point A to this line is simply the y-coordinate of A, which is 0, and the perpendicular distance from point B is the x-coordinate of B, which is also 0.

Therefore, the sum of the perpendicular distances from A and B on the tangent to the circle at the origin is 0+0+0

 

Posted by

HARSH KANKARIA

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