If the line 3x+2λy+12=0 is a diameter of the circle x2+y2−4x+6y+12=0, then the value of λ is
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Solution
Given line is 3x+2λy+12=0 Equation of the circle is x2+y2−4x+6y+12=0, Centre of the circle is C=(−g,−f)=(2,−3) Centre should lie on the equation of diameter, so 3(2)+2λ(−3)+12=0∴λ=3