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How do you find equations of lines, the equation of a circle, and the tangent to a circle?

Use coordinate geometry: find equations of straight lines including parallel and perpendicular lines, use the equation of a circle, and find the equation of a tangent to a circle.

A CCEA GCSE Further Mathematics answer on coordinate geometry, covering the gradient and equation of a line, parallel and perpendicular conditions, the distance and midpoint, the equation of a circle, and the tangent to a circle in the compulsory Pure unit.

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  1. What this dot point is asking
  2. Lines, gradients and intercepts
  3. Parallel and perpendicular lines
  4. The equation of a circle
  5. The tangent to a circle
  6. Why this matters

What this dot point is asking

Coordinate geometry studies lines and circles using algebra on the xyxy-plane, and CCEA GCSE Further Mathematics adds the circle to the straight-line work. You must find gradients and equations of lines, use the parallel and perpendicular conditions, compute distances and midpoints, recognise and use the equation of a circle, and find the tangent to a circle at a point. The link between a radius and a tangent is the signature idea of this topic.

Lines, gradients and intercepts

The gradient measures steepness as the change in yy over the change in xx between two points. Once you know a gradient and a point, the point-gradient form y−y1=m(x−x1)y - y_1 = m(x - x_1) gives the equation directly, which you then tidy into y=mx+cy = mx + c form.

The midpoint of two points is the average of the coordinates, (x1+x22,y1+y22)\left(\tfrac{x_1 + x_2}{2}, \tfrac{y_1 + y_2}{2}\right), and the distance between them comes from Pythagoras applied to the horizontal and vertical gaps.

Parallel and perpendicular lines

Two lines are parallel when their gradients are equal, so they never meet. Two lines are perpendicular when their gradients multiply to −1-1, which means each gradient is the negative reciprocal of the other.

The equation of a circle

A circle is the set of points a fixed distance (the radius) from a centre. Applying the distance formula gives its equation. Centred at the origin it is x2+y2=r2x^2 + y^2 = r^2; centred at (a,b)(a, b) it is (x−a)2+(y−b)2=r2(x - a)^2 + (y - b)^2 = r^2. You read the centre and radius straight from this form, so completing the square is sometimes needed to put a circle into it.

The tangent to a circle

A tangent touches the circle at exactly one point and is perpendicular to the radius drawn to that point. This perpendicular relationship is what lets you find its equation.

Why this matters

Coordinate geometry connects algebra and shape and supports several other topics: the line-and-circle intersection uses the simultaneous-quadratic method from quadratic theory; the perpendicular condition reappears in the tangent and normal work of calculus; and the distance formula relies on surds for exact answers. It also underpins the vector and force geometry of the Mechanics unit, so secure technique here pays off widely.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA Unit 1 (style)4 marksFind the equation of the line perpendicular to y=2x+5y = 2x + 5 that passes through the point (4,1)(4, 1).
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The gradient of the given line is 22, so the perpendicular gradient is the negative reciprocal, −12-\tfrac{1}{2}.

Use y−y1=m(x−x1)y - y_1 = m(x - x_1) with m=−12m = -\tfrac{1}{2} and the point (4,1)(4, 1): y−1=−12(x−4)y - 1 = -\tfrac{1}{2}(x - 4).

Expand: y−1=−12x+2y - 1 = -\tfrac{1}{2}x + 2, so y=−12x+3y = -\tfrac{1}{2}x + 3.

Marks are for the perpendicular gradient, substituting the point, and the final equation. The common error is keeping the gradient 22 or only flipping the sign without taking the reciprocal.

CCEA Unit 1 (style)5 marksA circle has equation x2+y2=25x^2 + y^2 = 25. Find the equation of the tangent to the circle at the point (3,4)(3, 4).
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The radius to (3,4)(3, 4) has gradient 4−03−0=43\dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}.

The tangent is perpendicular to the radius, so its gradient is the negative reciprocal, −34-\tfrac{3}{4}.

Use y−y1=m(x−x1)y - y_1 = m(x - x_1) at (3,4)(3, 4): y−4=−34(x−3)y - 4 = -\tfrac{3}{4}(x - 3), which expands to y=−34x+254y = -\tfrac{3}{4}x + \tfrac{25}{4}.

Marks are for the radius gradient, the perpendicular tangent gradient, and the equation. Forgetting that the tangent is perpendicular to the radius is the key error.

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