MATHS

Asked by SophiéRearrange the equation for the straight line to the form: ax+b=y and substitute this result for y into the equation for the circle (x-c)^2+(y-d)^2=e and you can rearrange to get a quadratic in x. Solve the quadratic. Since the line is a tangent to the circle, there will only be one repeated solution, which gives the x coordinate you are after. Substitute back into the equation of the line to find your y value.

The equation of the circle is: (x-a)^2 + (y-b)^2 = r^2 The equation of the tangent (linear ofc) is: y=mx+c At the point of contact for these lines, these functions will share the same co-ordinate x & y values. Therefore, you can substitute the value of the tangent (mx+c) into the value of y in the circle equation. This gives: (x-a)^2 + ((mx+c)-b)^2 = r^2 If you simplify the above, you can find 2 values for x. You can plug both values of x into the equation for the tangent to produce two y values. You now have 2 pairs of (x,y) co-ordinates. From a graphical sketch of the circle and the tangent, you can deduce which of the two co-ordinates is valid.

Hi A point of contact between a tangent and a circle is the only point touching the circle by this line, The point can be found either by : equating the equations; The line : y = mx +c The circle : (x-a)^2 + (y_b)^2 = r^2 The result will be the value of {x}which can be substituted in the equation of the line to find the value of the {y}. Or; by substituting the equation of the line in the equation of the circle to find the value of the {x} and then substitute this value again in the equation of the line to find the value of the {y}. Hope this will help.

It really depends on the information you are given. If you have the equations of both lines, you can do some algebra. Equations of circles are in the form (x-a)^2 + (y-b)^2 = r^2 And straight lines are y=mx+c With these you can just substitute the second equation into the first by replacing all the y's with mx+c. After you simplify, you should get a quadratic expression and solving that will give you an x coordinate. Using that you plug it back into y=mx+c to get the y coordinate. There are other ways to do this which require different peices of information. If you tell me what is given in the question, then I can tell you a more suitable method

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.

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.