Rearrange 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... more

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Dan Ritchie
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207 students helped

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3 months ago

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 fin... more

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Ali Rajani
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124 students helped

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3 months ago

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 equati... more

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Alaa Renno
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166 students helped

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4 months ago

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 giv... more

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Awais Naveed
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451 students helped

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4 months ago

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... more

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Will Wilson
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100 students helped

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4 months ago

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 jus... more

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Nick Gray
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264 students helped

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7 months ago

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.

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fowzi Begum
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231 students helped

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a year ago

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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Talha Ghannam
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65 students helped

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a year ago


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