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  • The extremities of a diameter of a circle are (1, 2) and (3, 4). Determine the equations of tangents which are parallel to this diameter.  Option:

The extremities of a diameter of a circle are (1, 2) and (3, 4). Determine the equations of tangents which are parallel to this diameter.

 

Option: 1

\mathrm{y=x-3 \quad \& \quad y=x-1}


Option: 2

\mathrm{y=x+3\: \&\: y=x-1}


Option: 3

\mathrm{y=2 x+3\: \&\: y=2 x-1}


Option: 4

\mathrm{y=x+3 \quad \& \quad y=x+1}


Answers (1)

best_answer

The equation of the circle is \mathrm{(x-1)(x-3)+(y-2)(y-4)=0} or \mathrm{x^2+y^2-4 x-6 y+11=0}

Centre is \mathrm{(2,3) r=\sqrt{2^2+3^2-11}=\sqrt{2}}

Slope of the diameter joining (1,2) and (3,4) is \mathrm{\frac{(4-2)}{(3-1)}}=1

Hence any line parallel to this diameter is   y = x + c          or     x – y + c = 0

If it is a tangent, then perpendicular from centre (2,3) is equal to the radius

\mathrm{\therefore \frac{2-3+c}{ \pm \sqrt{2}}=\sqrt{2} \quad \Rightarrow c-1= \pm 2, \quad \therefore c=3,-1}

Hence the tangents are y = x + 3, and y = x – 1

 

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Pankaj

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