A circle has equation x^2+y^2-8x+10y+41=0. A point on the circle has coordinates (8,-3). Find the equation of the tangent to the circle passing through this point.
From the equation x2+y2-8x+10y+41=0, we can find the centre of the circle by the middle two terms -8x+10y and multiplying them by -1/2. So the coordinates of the circle centre are (4,-5).Now we can find the gradient of the line passing through the centre of the circle and the point (8,-3) specified in the question. We do this using m=y2-y1/x2-x1=-3-(-5)/8-4=1/2. So the gradient of the line from the centre to the point 1/2. But we need the gradient of the tangent passing through this point. We can find this by switching the numerator and the denominator of the centre gradient and multiplying by -1. So the tangent gradient is 2/1*(-1)=-2.Now if we input the gradient into the standard line equation y=mx+c, we can see that y=-2x+c. We're still missing the constant c. To find c, we need to find a point that we know definitely lies on the tangent line. Why not the one we're already given (8,-3)! I recommend rearranging the line equation before plugging any numbers in, so rearrange for c to give c=y-mx. Input the coordinates of the point as well as the gradient to find c=-3-(-2)*8=-3+16=13.So the equation of the tangent line is y=-2x+13.
