πŸ’― Maths

How do I find the point of contact of a tangent to a curve or circle?

4 answers
Answered Jun 20Maths
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Will WilsonExperienced Tutor | Undergraduate in Mathematics and Philosophy 62 students helped

Let's answer each one of these parts in turn: Firstly how to find the tangent of a curve. Most of the curves you will see through GCSE and A-Level are polynomials. To get a tangent line in the standard form of y=mx+c (where m is the gradient and c is the y-intercept) we first need to find the gradient. We can either differentiate the curve (very basically you time the coefficient of each term with the power of x and take one off the power) and the substitute the tangent point into the equation or you can draw a tangent line from the point and calculate the gradient using 2 points on the line. The gradient is the difference in the y co-ordinates/ difference in the x co-ordinates. Then to calculate the y-intercept we can substitute the tangent point in the equation y=mx+c with your recently calculated gradient m. Now calculating tangents to circle. The equation for a tangent to a circle is y-b=m(x-a). We can use the methods above to calculate the gradient and (a,b) is the centre of the circle. I hope this helps.

Answered Mar 20Maths
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Nick GrayResourceful tutor and undergraduate at The University of Auckland237 students helped

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 & β€œnon-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.

Answered Jul 19Maths
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faizah BegumHiiii, I want to get you amazing gcse results204 students helped

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.

Answered Apr 19Maths
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Talha GhannamExperienced tutor with proven results in maths and economics at all levels!38 students helped

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.