If the lines `x+y=6 and x+2y=4`
are diameters of the circle which passes through the point (2, 6), then find
its equation.
Text Solution
Verified by Experts
Here center will be the point of intersection of the diameter, i.e., `C (8,-2)`. Also, the circle passes through the point P(2,6). Then radius is `CP=10`. Hence, the required equation is `(x-8)^(2)+(y+2)^(2)=10^(2)` or `x^(2)+y^(2)-16x+4y-32=0`
x+y=6 and x+2y=4 are two diameters of a circle. If the circel passes through the point (6,2), then its radius is-
3x + y = 5 and x+y+1 = 0 are two diameters to the circle which passes through the point (-2, 2). Find its equation. Also find the radius of the circle.
3x+y=5 and x+y-1=0 are the equations of two diamters to the circle,which passes through the point (-2,2).Find the equation of the circle.
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Show that the circle x^(2) + y^(2) + 4(x+y) + 4 = 0 touches the coordinates axes. Also find the equation of the circle which passes through the common points of intersection of the above circle and the straight line x+y+2 = 0 and which also passes through the origin.
