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  • Equation of the tangent to the circle, at the point (1,-1), whose centre is the point of intersection of the straight lines <img alt="\mathrm{x-y=1 \, \, and \, \, 2 x+y=3}" src="https://entranceco

Equation of the tangent to the circle, at the point (1,-1), whose centre is the point of intersection of the straight lines \mathrm{x-y=1 \, \, and \, \, 2 x+y=3} is

Option: 1

x+4 y+3=0


Option: 2

3 x-y-4=0


Option: 3

x-3 y-4=0


Option: 4

4 x+y-3=0


Answers (1)

best_answer

Point of intersection of lines

x-y=1 \, \, and\, \, 2 x+y=3\, \, is \, \, O\left(\frac{4}{3}, \frac{1}{3}\right)

Slope of O P=\frac{\frac{1}{3}+1}{\frac{4}{3}-1}=\frac{\frac{4}{3}}{\frac{1}{3}}=4$$\therefore \quad

Slope of tangent =-\frac{1}{4}

So, equation of tangent at P(1,-1) is

y+1=-\frac{1}{4}(x-1) \Rightarrow 4 y+4=-x+1 \Rightarrow x+4 y+3=0

Posted by

Devendra Khairwa

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