What is the equation of the tangent to the circle (x-5)^2+(y-3)^2=9 at the points of intersection of the circle with the line 2x-y-1=0
Teh points of intersection can be found by subbing in y=2x-1 to the equation of the circle. (x-5)2+(2x-1-3)2=5x2-26x+33=9 so we have 5x2-26x+24=0. Solving the quadratic equation gives x1=4 and x2=1.2. If x=4 we have two corresponding points on the circle: (4-5)2+(y-3)2=9 giving y1=3+2sqrt(2) or y2=3-2sqrt(2). The equation of the corresponding tangent can be found in various ways for instance by implicit differentiation of the equation of the circle: 2x-10+2ydy/dx-6dy/dx=0 so we have dy/dx=(2x-10)/(6-2y) is the slope of the tangent. At the first intersection point: subbing in x=4 and y=3+2sqrt(2) gives slope=sqrt(2)/4. So the equation of the tangent line is sqrt(2)/4*(x-4)+y-3-2sqrt(2)=0. At the second intersection point x=4 and y=3-2sqrt(2) gives slope=-sqrt(2)/4. So the equation of the tangent line is -sqrt(2)/4*(x-4)+y-3-2*sqrt(2)=0. We have one last case left to examine if x=1.2, but subbing this in to the equation of the circle there is no sufficent y.
