Test Index

Circle

Section: Mathematics

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Question : 85 of 89

Marks: +1, -0

If

Q(h, k)

is the inverse point of the point

P(1,2)

with respect to the circle

x^2+y^2-4x+1=0

2h + k =

[AP EAPCET 18 May 2024 Shift 1]

3
4
7
11

Solution:

The equation of pole with respect ts the point

\beta, 2

to the given circle is

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

\;\Rightarrow -x+2y-1=0 \;\text{or}\; x-2y+1=0

Inverse of point

P(1,2)

is the foot

(h, k)

of the perpendicalar from the point

(1,2)

to the line

x-2y+1=0

\;\Rightarrow\;\;\;\frac{h-1}{1}=\;\frac{k-2}{-2}=-\left[\;\frac{1-4+1}{1^2+(-2^{3.})}\right]

\;\Rightarrow\;\; h-1 = \;\frac{K-2}{-2} = \;\frac{2}{5}

\;\Rightarrow\;\; h-1 = \;\frac{2}{5} \;\text{and} \;\; \frac{K-2}{-2} = \;\frac{2}{5}

\;\Rightarrow\;\; h = \;\frac{7}{5} \;\text{and} \; K = \;\frac{-4}{5} + 2 = \;\frac{6}{5}

Hence,

2h + K = \;\frac{14}{5} + \;\frac{6}{5} = \;\frac{20}{5} = 4

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