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  • Please help! 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 :

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 :

  • Option 1)

    4x  + y − 3 = 0

  • Option 2)

    x + 4y + 3 = 0

  • Option 3)

    3x − y − 4 = 0

  • Option 4)

     x − 3y − 4 = 0

     

 

Answers (1)

As we learnt in 

Condition for perpendicular lines -

m_{1}m_{2}= -1

- wherein

Here m_{1},m_{2}  are the slope of perpendicular lines.

 Centre of  circle is \left ( \frac{4}{3},\frac{1}{3} \right )

slpoe of OP = \frac{-1-\frac{1}{3}}{1-\frac{4}{3}} =4

Since tangent and radius are perpendicular

So slope of tangent will be   \frac{-1}{4}

equation \Rightarrow \frac{y+1}{x-1} = \frac{-1}{4}

4y+x=-3

\Rightarrow x+4y+3=0


Option 1)

4x  + y − 3 = 0

Incorrect

Option 2)

x + 4y + 3 = 0

Correct

Option 3)

3x − y − 4 = 0

Incorrect

Option 4)

 x − 3y − 4 = 0

 

Incorrect

Posted by

Sabhrant Ambastha

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