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Circle

Section: Mathematics

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Question : 15 of 89

Marks: +1, -0

If

Q

is the inverse point of

P(-1,1)

with respect to the circle

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

, then the line containing

Q

[AP EAPCET 22 May 2025 S1]

$x-3y-2=0$
$x-y+1=0$
$x+y-2=0$
$2x-3y+5=0$

Solution:

Here

Q

is the inverse point of

P(-1,1)

with respect to the circle

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

\Rightarrow (x-1)^2+(y+1)^2=2

Centre of circle

O \equiv (1,-1)

\therefore O P \cdot O Q=r^2

\Rightarrow O Q = \frac{r^2}{O P} = \frac{2}{\sqrt{(1+1)^2+(-1-1)^2}}

= \frac{2}{\sqrt{8}} = \frac{1}{\sqrt{2}}

O P, O Q

are collinear

\therefore

Slope of

O P=-1

Q

point lie on

O P Q

\therefore \frac{x-1}{\cos \theta} = \frac{y+1}{\sin \theta} = O Q

\Rightarrow \frac{x-1}{\sqrt{2}} = \frac{y+1}{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}}

x-1 = \frac{-1}{2}, \quad y+1 = \frac{1}{2}

x = \frac{1}{2}, y = \frac{-1}{2}

\therefore Q \equiv \left( \frac{1}{2}, \frac{-1}{2} \right) \text{ lie on line }

x-3y-2=0

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