posted by Kelly .
verify that the point (3,2) lies on the circle x^2+y^2-8x+2y+7=0, and find the equation of the tangent at this point.
find the length of the tangents from the point (5,7) to the circle x^2+y^2-4x-6y+9=0
find the equation of the common chord of the circles x^2+y^2-6x+2y+1=0 and x^2+y^2+4x-2y-13=0
#1. Have you tried plugging x=3 and y=2 into the equation? Do it and verify that the equation is satisfied there.
Then get the circle's slope dy/dx at (3,2) by implicit differentiation.
Once you know one point on the line (3,2), and the slope at that point, the equation of the line can easily be written.
(y-2) = (x-3)*dy/dx
#2 Rewrite the circle equation
(x-2)^2 -4 + (y-3)^2 -9 + 9 = 0
(x-2)^2 + (y-3)^2 = 4
That tells you the circle's center location and radius. Plot it and the point (5,7) on a graph. Draw a line from (5,7) to the circle's center at (2,3). From the length of that line (a hypotenuse) and the radius of the circle, use the Pythagorean theorem to get the length of the tangent. A tangents, radius to the tangency point, and hypotenuse form a right triangle.
3. That's all I have time for. Please show your own work on future posts