If a circle S passing through the points A(1,2) and B(2,1) has its centre C located in the third quadrant at a distance of 7 2 units from AB , then the point P (1,-2)

Step-by-step solution

Now, aligned & AP = 1 2 1^2+(-1)^2 = 1 2 & and PC = 7 2 & so AC = ( 1 2 )^2+ ( 7 2 )^2 =5 aligned Now, AC = BC AC ^2= BC ^2 aligned & ( h -1)^2+( k -2)^2=( h -2)^2+( k -1)^2 & h = k ... (i) aligned aligned & And A C^2=5^2 (h-1)^2+(h-2)^2=25 & h^2+1-2 h+h^2+4-4 h=25 & 2 h^2-6 h-20 h^2-3 h-10=0 & (h-5)(h+2)=0 h=-2,5 aligned So (h, k)=(-2,-2) Distance between (1-2) &(-2,-2)=3 and 3 < 5 So the point P (1,-2) lies inside the circle.