=[(1)/(2)*(x)/(2)+y=0.8],[(7)/(x+(y)/(2))=10]

Solve the following system of equations: (x)/(2)+y=0.8,(7)/(x+(y)/(2))=10

solve using ellimination method (x)/(2)+y=0.8 and (7)/(x)+(y)/(2)=10

Solve the following system of equations: x/2+y=0. 8 ,\ \ 7/(x+y/2)=10

The circles whose equations are x^(2)+y^(2)+10x-2y+22=0 and x^(2)+y^(2)+2x-8y+8=0 touch each other. The circle which touch both circles at, the point of contact and passing through (0,0) is, 1) 9(x^(2)+y^(2))-15x-20y=0, 2) 5(x^(2)+y^(2))-18x-80y=0 3) 7(x^(2)+y^(2))-18x-80y=0, 4) x^(2)+y^(2)-9x-40y=0 ]]

The radical centre of circles represented by S_(1)-=x^(2)+y^(2)-7x-6y-4=0,S_(2)-=x^(2)+y^(2)+10x+6y-4=0

Solve (x+2)/(y+2)+2=0,(x-4)/(y-2)=(x-1)/(y+7)

The locus of a point "P" ,if the join of the points (2,3) and (-1,5) subtends right angle at "P" is x^(2)+y^(2)-x-8y+13=0 x^(2)-y^(2)-x+8y+3=0 x^(2)+y^(2)-4x-4y=0,(x,y)!=(0,4)&(4,0) x^(2)+y^(2)-x-8y+13=0,(x,y)!=(2,3)&(-1,5)

If (x+y)/(0.01)=7 , then (1)/(2x+2y)=

Find the equation of the common tangent of the following circles at their point of contact x^(2)+y^(2)=10x-2y+22=0, x^(2)+y^(2)+2x=8y+8=0