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Q.
The equation of the diameter of the circle $ {{x}^{2}}+{{y}^{2}}+2\,x-4\,y-11=0 $ which bisects the chords intercepted on the line $ 2\,x-y+3=0 $ is:
Bihar CECEBihar CECE 2001
Solution:
The perpendicular distance from centre to the chord its bisects the chord.
Given equation of circle is
$x^{2}+y^{2}+2 \,x-4 \,y-11=0$
$\therefore $ Centre is, $(-1,2)$.
Since, the diameter of the circle is perpendicular to the chord of the circle. Hence, the equation of diameter is $x+2 \,y+\lambda=0 \ldots$..(i)
Its passes through the centre $(-1,2)$ of the circle
$\Rightarrow -1+4+\lambda=0$
$\Rightarrow \lambda=-3$
Putting the value of $\lambda$ in Eq. (i), we get
$x+2 \,y-3=0$