Lines Meeting Circles: Tangents and Chords

Mathematics · Geometry · ages 16-17

Name ______________________ Date ____________

The tangent to a circle at a point is at right angles to the radius drawn to that same point.
Circle one: True False
You want to find where y = 3x - 1 meets a circle. What is the first move?
- work out the radius of the circle
- plot both on squared paper and read the points off
- put 3x - 1 in place of y in the circle equation
- complete the square on the circle equation
What is the discriminant of 3x² - 4x + 2 = 0?
Answer: ______________
After substituting a line into a circle, the quadratic has discriminant 0. What does that say?
- the line misses the circle entirely
- the line touches the circle at exactly one point
- the line cuts the circle at two points
- the line passes through the centre
Substitute y = 2x into x² + y² = 45. What is the positive value of x?
Answer: ______________
Tom finds a tangent at a point on a circle by giving it the same gradient as the radius to that point. Where is the slip?
- the tangent gradient comes from the centre, not from the point
- the tangent gradient is the negative reciprocal of the radius gradient
- a tangent has no gradient, because it touches only once
- the radius gradient has to be doubled first
Find the tangent to the circle x² + y² = 100 at the point (8, 6).
- 3x + 4y - 48 = 0
- 4x - 3y - 14 = 0
- 4x + 3y - 50 = 0
- 3x + 4y - 50 = 0
Ella substitutes y = x + 2 into x² + y² = 10 and writes x² + x² + 4 = 10. Where is the slip?
- the 10 should have been squared as well
- she should have substituted for x rather than for y
- the circle equation needed completing the square first
- squaring (x + 2) gives x² + 4x + 4, so the 4x is missing
Find the tangent to the circle (x - 2)² + (y + 1)² = 169 at the point (14, 4).
- 12x + 5y - 188 = 0
- 5x + 12y - 118 = 0
- 12x - 5y - 148 = 0
- 12x + 5y - 168 = 0
The line 2x + y = 10 touches the circle x² + y² = 20 at one point. Which point?
- (2, 6)
- (5, 0)
- (0, 10)
- (4, 2)

Answer key

For grown-ups. Fold this page away before handing over the rest.

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True · That right angle is what makes the second route to a tangent work: the two gradients multiply to -1.
put 3x - 1 in place of y in the circle equation · Substituting the line into the circle turns two equations into one quadratic, and the quadratic carries the whole answer.
-8 · With a = 3, b = -4 and c = 2, b² - 4ac is 16 - 24, which is -8.
the line touches the circle at exactly one point · A discriminant of zero gives one repeated root, so there is one meeting point and the line is a tangent.
3 · Substituting gives x² + (2x)² = 45, which is x² + 4x² = 45, so 5x² = 45, x² = 9 and the positive x is 3.
the tangent gradient is the negative reciprocal of the radius gradient · The tangent meets the radius at a right angle, so the two gradients multiply to -1. Giving them the same gradient would draw a line along the radius, straight through the circle.
4x + 3y - 50 = 0 · The radius runs from (0, 0) to (8, 6), so its gradient is 6/8 = 3/4. The tangent gradient is -4/3, and y - 6 = -(4/3)(x - 8) gives 4x + 3y - 50 = 0.
squaring (x + 2) gives x² + 4x + 4, so the 4x is missing · A bracket squared is not each term squared. (x + 2)² is x² + 4x + 4, so Ella lost the middle term and with it one whole root.
12x + 5y - 188 = 0 · The centre is (2, -1), so the radius gradient is (4 - (-1)) / (14 - 2) = 5/12. The tangent gradient is -12/5, and y - 4 = -(12/5)(x - 14) gives 12x + 5y - 188 = 0.
(4, 2) · Rearranging gives y = 10 - 2x. Substituting gives 5x² - 40x + 80 = 0, which is x² - 8x + 16 = 0, or (x - 4)² = 0. So x = 4 and y = 10 - 8 = 2.