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Circle

Section: Mathematics

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Question : 7 of 89

Marks: +1, -0

If

2x-3y+1=0

is the equation of the polar of a point

P(x_1, y_1)

with respect to the circle

x^2+y^2-2x+4y+3=0

3x_1-y_1=

[AP EAPCET 24 May 2025 S1]

$\frac{1}{3}$
$-3$
$3$
$-\frac{1}{3}$

Solution:

Given circle is

x^2+y^2-2x+4y+3=0

So, center

=(-g,-f)=\left(\frac{-2}{2},\frac{4}{2}\right)

=(-1,2)\;\text{and}\;c=3

Also, the equation of the polar of a point

(x_1, y_1)

w.r.t the circle

x^2+y^2+2gx+2fy+c=0

is

x x_1+y y_1+g(x+x_1)+f(y+y_1)+c=0

\Rightarrow x x_1+y y_1-(x+x_1)+2(y+y_1)+3=0

\Rightarrow x x_1+y y_1-x-x_1+2y+2y_1+3=0

\Rightarrow (x_1-1)x+(y_1+2)y-x_1+2y_1+3=0\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

Given, polar equation is

2x-3y+1=0\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(ii)

By comparing the coefficients of two equations, we get

\frac{x_1-1}{2}=\frac{y_1+2}{-3}

=\frac{-x_1+2y_1+3}{1}

\Rightarrow\;\; \frac{x_1-1}{2}=\frac{y_1+2}{-3}\;\text{and}\;

\frac{x_1-1}{2}=\frac{-x_1+2y_1+3}{1}

\Rightarrow -3x_1+3\;=2y_1+4\;\text{and}\;x_1-1

=-2x_1+4y_1+6

\Rightarrow 3x_1+2y_1\;=-1\;\text{and}\;3x_1-4y_1-7=0

By solving these two equations, we get

x_1=\frac{5}{9}

and

y_1=\frac{-4}{3}

\;\text{Now,}\; 3x_1-y_1 = 3\left(\frac{5}{9}\right)-\left(\frac{-4}{3}\right)

=\frac{5}{3}+\frac{4}{3}=\frac{9}{3}=3

\therefore 3x_1-y_1=3

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