How Do You Know if a Line Is Tangent to a Circle
How do I find the bespeak of contact of a tangent to a curve or circumvolve?
Rearrange the equation for the straight line to the form: ax+b=y and substitute this consequence for y into the equation for the circle (x-c)^2+(y-d)^ii=east and you tin rearrange to get a quadratic in x. Solve the quadratic. Since the line is a tangent to the circle, there will just be one repeated solution, which gives the x coordinate you are subsequently. Substitute back into the equation of the line to notice your y value.
The equation of the circle is: (x-a)^2 + (y-b)^2 = r^two The equation of the tangent (linear ofc) is: y=mx+c At the betoken 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: (ten-a)^2 + ((mx+c)-b)^2 = r^2 If you simplify the above, you can find 2 values for 10. You can plug both values of x into the equation for the tangent to produce two y values. You at present take 2 pairs of (10,y) co-ordinates. From a graphical sketch of the circle and the tangent, you can deduce which of the two co-ordinates is valid.
Hullo A indicate of contact between a tangent and a circle is the simply point touching the circumvolve 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^ii The result volition be the value of {x}which can exist substituted in the equation of the line to find the value of the {y}. Or; past substituting the equation of the line in the equation of the circle to find the value of the {x} and and then substitute this value again in the equation of the line to find the value of the {y}. Hope this will assistance.
It really depends on the information you are given. If you have the equations of both lines, you can practise some algebra. Equations of circles are in the form (x-a)^two + (y-b)^2 = r^2 And straight lines are y=mx+c With these you can just substitute the second equation into the first past replacing all the y'southward with mx+c. After you simplify, you should become a quadratic expression and solving that volition requite you an x coordinate. Using that you plug it back into y=mx+c to get the y coordinate. There are other ways to practise this which require dissimilar peices of information. If you tell me what is given in the question, then I tin can tell y'all a more suitable method
Let's answer each 1 of these parts in turn: Firstly how to detect the tangent of a bend. About of the curves yous will see through GCSE and A-Level are polynomials. To become a tangent line in the standard form of y=mx+c (where m is the gradient and c is the y-intercept) nosotros first need to observe the gradient. Nosotros tin either differentiate the curve (very basically yous fourth dimension 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 tin can draw a tangent line from the point and summate the gradient using two points on the line. The slope is the difference in the y co-ordinates/ difference in the x co-ordinates. Then to summate the y-intercept we tin substitute the tangent signal in the equation y=mx+c with your recently calculated gradient m. At present calculating tangents to circle. The equation for a tangent to a circle is y-b=m(ten-a). We tin can utilise the methods above to summate the gradient and (a,b) is the centre of the circle. I hope this helps.
more specifically than above, merely post-obit through with the algebra: if the circle has an equation in the format: ( (x-a)^two + (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 offset simply use mx + c) to have a single quadratic equation in terms of x (afterwards expanding and combining 10^two, x & "not-x" terms). lastly apply 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 10 generically is as follows: (m^two + i)10^two - 2(a + b - c)10 + (a^2 + (b-c)^two) - r^2 = 0 follow instructions above + utilize 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 and then rearrange to notice the x value and substitute to observe y values you lot tin practice this eg/ by simultaneous equations.
Find the gradient of the line from the center of the circle to the indicate on the circle you need, and so utilize that to find the slope of the tangent. Finally, use y=mx+c with this gradient and the point on the circumvolve it touches.
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