How do I find the point of contact of a tangent to a curve or circle?
more specifically than above, but following through with the algebra: if the circle has an equation in the format: ( (x-a)^2 + (y-b)^2 = r^2 ........... (1) [circle centred at coordinates (a,b) with radius r] and the tangent (or line) has equation in the format: y = mx + c ...... (2) [line with slope m, intercept c] can substitute the second equation into the first (where y appears in first just use mx + c) to have a single quadratic equation in terms of x (after expanding and combining x^2, x & “non-x” terms). lastly use quadratic formulae to solve for 2 potential solutions & substitute each solution back into the line equation ( y = mx + c) to solve for the corresponding y coordinate. solution for x generically is as follows: (m^2 + 1)x^2 - 2(a + b - c)x + (a^2 + (b-c)^2) - r^2 = 0 follow instructions above + use quadratic formulae to complete.
the point of contact = points of intersections to find this you can equate the both equations of the circle and the tangent then rearrange to find the x value and substitute to find y values you can do this eg/ by simultaneous equations.
Find the gradient of the line from the centre of the circle to the point on the circle you need, then use that to find the gradient of the tangent. Finally, use y=mx+c with this gradient and the point on the circle it touches.
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