what is the equation of circle which pass thrugh the points A(-1,4), B(1,2) and which touch the line 3x-y-3=0????
24Eqn of circle with centre (x1,y1) and radius r
(x-x_1)^2+(y-y_1)^2=r^2
Now put those points A and B and get 2 eqns
Since circle touches line 3x-y-3=0
So perpendicular distance from centre =radius\left|\frac{3x_1-y_1-3}{\sqrt{3^2+1}} \right|=r
Now 3 eqns and 3 variables ...solve it[1]
106There's a shorter method.
Consider the chord AB. its midpoint is (0,3)
So centre is along the line (y-3) = 1(x-0) i.e. y=x+3
So, centre is (h,h+3) say. So r = ((h-1)2+(h+1)2)0.5 = √2h2+2
now square of perp. distance of tangent line from centre is equal to r2
=> (2h-6)2 = 10(2h2+2)
=> h2-6h+9 = 10h2+10
=> 9h2+6h+1=0
=> 3h+1=0
=> h=-1/3
now u can get the eqn of circle easily