AQA · Mathematics · 8300
GCSE Maths: Equation of a Tangent to a Circle
Algebra · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Equation of a Tangent to a Circle (Corbettmaths Video 372), and every card links back to the exact page it came from. Practice tests are drawn from this deck’s own cards. Source material © Corbettmaths — used with attribution.
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How is the gradient of a tangent to a circle related to the gradient of the radius at that point?
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The tangent is perpendicular to the radius, so its gradient is the negative reciprocal of the radius gradient (they multiply to −1).
For a circle x² + y² = r², what is the gradient of the radius to the point (a, b)?
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b/a — the radius joins the origin O to (a, b).
Outline the steps to find the equation of the tangent to x² + y² = r² at the point (a, b).
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1) Gradient of radius = b/a. 2) Tangent gradient = −a/b (negative reciprocal). 3) Substitute into y − b = m(x − a).
What is the equation of the tangent to x² + y² = 8 at the point (2, 2)?
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y = −x + 4. Radius gradient is 1, so the tangent gradient is −1.
What is the gradient of the tangent to x² + y² = 40 at the point (2, 6)?
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−1/3. The radius gradient is 6/2 = 3, and the tangent gradient is its negative reciprocal.
What is the gradient of the tangent to x² + y² = 29 at the point (−5, 2)?
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5/2. The radius gradient is 2/(−5) = −2/5, so the tangent gradient is the negative reciprocal, 5/2.
What is the equation of the tangent to x² + y² = 68 at the point (2, 8)?
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x + 4y = 34 (equivalently y = −¼x + 8½). Radius gradient 4 → tangent gradient −1/4.
How do you find the centre of a circle given the endpoints of a diameter?
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Take the midpoint of the two endpoints.
What is the radius of the circle with diameter from D(−12, −5) to E(12, 5)?
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13. The centre is (0, 0) and the radius is √(12² + 5²) = √169 = 13.
The circle with centre (0, 0) and radius 13 has equation x² + y² = ?.
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169
Why do parallel tangents to a circle x² + y² = r² touch it at diametrically opposite points?
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Parallel tangents have the same gradient, so their radii have the same gradient too — the radii lie on one line through the centre, giving points (a, b) and (−a, −b).
How do you find where a tangent line crosses the x-axis?
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Set y = 0 in the tangent's equation and solve for x.
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