Question

If (1,2) (2,a) are extremities of a diameter
of the circle x² + y2 – 3x – 4y+6=0 then a=​

Answer 1

Step-by-step explanation:

x^2-12x+y^2+4y+6=0\\\\(x-6)^2-6^2+(y+2)^2-2^2+6=0\\\\(x-6)^2+(y+2)^2=34\\\\Center=(6,-2)\\\\Let\ equation\ of\ Diameter\ be: y=mx+c\\\\(6,-2)\ is\ on\ the\ line\\\\-2=m\ 6+c,\ \ =>\ c=-6m-2\\\\Equation\ of\ diameter=y=mx-6m-2\\

This is a parametric equation for the diameter of the circle. Given a value of m, you get the equation of diameter with that slope.