This Year 11 equation of a circle and tangents revision notes is written for pupils aged 15-16 and follows the National Curriculum. In short, it shows you how to use x^2 + y^2 = r^2 and find tangent equations. Before you start, it helps to revise functions and graph transformations and rates of change and areas under graphs, because this lesson builds on them.
Year 11 equation of a circle and tangents revision notes: what you will learn
This whole topic is Higher tier. It joins coordinate geometry, Pythagoras and perpendicular lines. At GCSE the circle always has its centre at the origin \((0,0)\).
Starter activity
Draw a circle of radius 5 on squared paper with its centre at the origin. Mark every point with whole-number coordinates that lies on the circle, such as \((3,4)\). Check that \(x^2+y^2=25\) for each one. Now draw the radius to \((3,4)\) and use a set square to draw the line at right angles to it at that point. What do you notice about this line and the circle?
The equation of a circle
Theorem (Higher): equation of a circle
Take any point \((x,y)\) on a circle of radius \(r\) with centre \((0,0)\). The horizontal distance from the centre is \(x\) and the vertical distance is \(y\). By Pythagoras, \(x^2+y^2=r^2\): \[x^2+y^2=r^2\] The number on the right is the radius squared. The circle \(x^2+y^2=25\) has radius 5. The circle \(x^2+y^2=10\) has radius \(\sqrt{10}\).
Definition (Higher): points on, inside and outside
A point can be on, inside or outside a circle. Substitute the coordinates into \(x^2+y^2\) and compare with \(r^2\).
- If \(x^2+y^2=r^2\) the point is on the circle.
- If \(x^2+y^2\lt r^2\) the point is inside.
- If \(x^2+y^2>r^2\) the point is outside.
Worked example 1: finding the equation
Find the equation of the circle, centre the origin, passing through \((-5,12)\).
\(r^2=(-5)^2+12^2=25+144=169\). The equation is \(x^2+y^2=169\), so the radius is 13.
Tangents
Definition: tangent
A tangent is a straight line that touches the circle at exactly one point.
Rule: tangent and radius
A tangent is perpendicular to the radius at the point of contact. The gradients multiply to give \(-1\): if the radius has gradient \(m\), the tangent has gradient \(-\dfrac1m\).
Method: finding the equation of a tangent
- Check that the point lies on the circle.
- Find the gradient of the radius from \((0,0)\) to the point: \(\dfrac{y}{x}\).
- Find the perpendicular gradient: negative reciprocal.
- Use \(y-y_1=m(x-x_1)\) and tidy up.
Worked example 2: tangent at \((3,4)\)
Find the tangent to \(x^2+y^2=25\) at \((3,4)\).
Check: \(9+16=25\). Radius gradient \(=\dfrac43\). Tangent gradient \(=-\dfrac34\). \(y-4=-\dfrac34(x-3)\Rightarrow4y-16=-3x+9\Rightarrow3x+4y=25\).
Worked example 3: tangent at \((-6,8)\)
Find the tangent to \(x^2+y^2=100\) at \((-6,8)\).
Radius gradient \(=\dfrac{8}{-6}=-\dfrac43\). Tangent gradient \(=\dfrac34\). \(y-8=\dfrac34(x+6)\Rightarrow4y-32=3x+18\Rightarrow3x-4y=-50\), or \(y=\dfrac34x+12.5\).
Worked example 4: tangent at \((4,2)\)
Find the tangent to \(x^2+y^2=20\) at \((4,2)\) and where it crosses the \(x\)-axis.
Check: \(16+4=20\), so the point is on the circle. The radius gradient is \(\dfrac24=\dfrac12\), so the tangent gradient is \(-2\). Then \(y-2=-2(x-4)\), which gives \(y=-2x+10\). It crosses the \(x\)-axis where \(0=-2x+10\), so at \(x=5\).
Watch out: points on an axis
If the point is on an axis, the radius is horizontal or vertical, so the tangent is vertical or horizontal. For \(x^2+y^2=16\), the tangent at \((0,4)\) is the line \(y=4\) and the tangent at \((-4,0)\) is the line \(x=-4\). Do not use \(-\dfrac1m\) here because a vertical line has no gradient.
Key fact (Higher): why the method works
The radius to the point \((a,b)\) has gradient \(\dfrac{b}{a}\). The tangent meets it at \(90^\circ\), so it has gradient \(-\dfrac ab\). Putting this into \(y-b=-\dfrac ab(x-a)\) and tidying gives \(ax+by=a^2+b^2=r^2\). For \((3,4)\) on \(x^2+y^2=25\) this gives \(3x+4y=25\). Use this quick form to check your answer, but show the gradient method in the exam to earn the marks.
Where a line meets a circle
Rule (Higher): line and circle
To find where a line crosses a circle, substitute the line into the circle equation and solve the quadratic.
- Two solutions: the line cuts the circle at two points.
- One (repeated) solution: the line is a tangent.
- No real solutions: the line misses the circle.
Worked example 5: a line meeting a circle
Find where \(y=x+1\) meets \(x^2+y^2=25\).
\(x^2+(x+1)^2=25\Rightarrow2x^2+2x-24=0\Rightarrow x^2+x-12=0\Rightarrow(x+4)(x-3)=0\). If \(x=3\), \(y=4\). If \(x=-4\), \(y=-3\). The points are \((3,4)\) and \((-4,-3)\).
Reasoning questions
Worked example 6: showing a tangent and tangent length
Show that \(y=5\) is a tangent to \(x^2+y^2=25\). Find the tangent length from \(P=(5,0)\) to the circle \(x^2+y^2=20\).
Substitute \(y=5\): \(x^2+25=25\), so \(x^2=0\) and there is only one solution, \(x=0\). One point of contact means a tangent, touching at \((0,5)\). State the conclusion in words. For the tangent length, triangle \(OTP\) has a right angle at \(T\), so \(PT^2=OP^2-r^2=25-20=5\) and \(PT=\sqrt5\).
Watch out: common mistakes
- Writing the radius instead of the radius squared: \(x^2+y^2=5\) is not a circle of radius 5.
- Using the radius gradient as the tangent gradient. Take the negative reciprocal.
- Forgetting to flip and change the sign: the perpendicular of \(\dfrac43\) is \(-\dfrac34\), not \(\dfrac34\) and not \(-\dfrac43\).
- Not checking that the point is actually on the circle.
Worked example 7: Finding the centre by completing the square
Find the centre and radius of the circle \(x^2+y^2-4x+6y-12=0\).
Group the terms: \((x^2-4x)+(y^2+6y)=12\). Complete each square: \((x-2)^2-4+(y+3)^2-9=12\). Rearrange: \((x-2)^2+(y+3)^2=25\). The centre is \((2,-3)\) and the radius is 5. Always add the numbers you subtracted to the right-hand side.
Worked example 8: Perpendicular gradients
A radius has gradient \(-\dfrac25\). What is the gradient of the tangent at the end of the radius?
The tangent is perpendicular to the radius, so its gradient is the negative reciprocal: flip the fraction and change the sign. \(-\dfrac25\) becomes \(\dfrac52\). Check: \(-\dfrac25\times\dfrac52=-1\). If the radius is horizontal, the tangent is vertical, and vice versa.
Worked example 9: Finding where a line is a tangent
Show that the line \(y=3\) is a tangent to the circle \((x-1)^2+(y-1)^2=4\) and find the point of contact.
Substitute \(y=3\): \((x-1)^2+4=4\), so \((x-1)^2=0\) and \(x=1\). There is exactly one solution, so the line touches the circle at one point: \((1,3)\). This is the point directly above the centre \((1,1)\) at a distance 2, which is the radius.
Worked example 10: A tangent through a point on the circle with a negative gradient
Find the tangent at \((4,-3)\) to the circle \(x^2+y^2=25\).
Gradient of the radius \(=\dfrac{-3}{4}\). Tangent gradient \(=\dfrac43\). \(y+3=\dfrac43(x-4)\), so \(3y+9=4x-16\) and \(4x-3y=25\). Check \((4,-3)\): \(16+9=25\) ✓. The coefficients are the coordinates of the point, which gives the shortcut \(4x-3y=25\) for the circle \(x^2+y^2=25\) at \((4,-3)\).
Worked example 11: The length of a tangent
A tangent is drawn from the point \((0,13)\) to the circle \(x^2+y^2=25\). Find its length.
The distance from the centre to the external point is 13. The radius to the point of contact is perpendicular to the tangent, forming a right-angled triangle with hypotenuse 13 and one side 5. The tangent length is \(\sqrt{13^2-5^2}=\sqrt{144}=12\).
Worked example 12: Intersection of two curves
Find where the circle \(x^2+y^2=20\) meets the line \(y=2x\).
Substitute: \(x^2+4x^2=20\), so \(5x^2=20\) and \(x^2=4\), giving \(x=\pm2\). The points are \((2,4)\) and \((-2,-4)\). They are opposite ends of a diameter because the line passes through the centre.
Worked example 13: Equation of a circle from three pieces of information
A circle has centre \((2,-1)\) and passes through \((5,3)\). Find its equation and decide whether \((2,4)\) is inside the circle.
\(r^2=(5-2)^2+(3+1)^2=9+16=25\), so the equation is \((x-2)^2+(y+1)^2=25\). For \((2,4)\): \((0)^2+(5)^2=25\), which equals \(r^2\), so the point is on the circle, not inside.
Worked example 14: A full tangent question with a diagram
A circle has centre \(O(0,0)\) and radius \(\sqrt{40}\). The point \(A(6,2)\) is on the circle. Find the equation of the tangent at \(A\) and the point where it crosses the \(y\)-axis.
Check: \(6^2+2^2=40\) ✓. The gradient of \(OA\) is \(\dfrac26=\dfrac13\), so the tangent gradient is \(-3\). Using \(y-2=-3(x-6)\) gives \(y=-3x+20\). Where \(x=0\), \(y=20\), so the tangent crosses the \(y\)-axis at \((0,20)\). Notice the shortcut for a circle centred at the origin: the tangent at \((a,b)\) is \(ax+by=r^2\), here \(6x+2y=40\), which simplifies to \(3x+y=20\). Both forms agree.
Note: three standard question types
Questions on this topic nearly always ask one of three things: (1) write or read the equation of a circle, (2) find the equation of a tangent at a point, using perpendicular gradients, or (3) find where a line meets a circle by substituting and solving a quadratic. For type 3, the discriminant tells you how many intersections there are: \(b^2-4ac>0\) means two points, \(=0\) means a tangent, and \(\lt 0\) means no intersection.
Watch out: sign and radius errors
- The equation \((x-3)^2+(y+2)^2=25\) has centre \((3,-2)\), not \((-3,2)\). The signs in the brackets are the opposite of the centre.
- The right-hand side is \(r^2\), so the radius of \(x^2+y^2=49\) is 7, not 49.
- The tangent gradient is the negative reciprocal, not just the negative.
- When the radius is horizontal or vertical, use \(x=\ldots\) or \(y=\ldots\) for the tangent; do not divide by zero.
Did you know?
The equation of a circle is a direct use of Pythagoras\' theorem: a point \((x,y)\) is at distance \(r\) from the centre exactly when \(x^2+y^2=r^2\). The French mathematician and philosopher René Descartes, who lived in the 17th century, was the first to describe shapes with equations like this by putting them on a grid. That is why the grid you use in school is called a Cartesian grid.
Otto’s tip
Draw a quick sketch with the circle, the point and a rough tangent. A tangent at a point in the top right has a negative gradient, so if your answer is positive, something has gone wrong. Check the equation by substituting the point back in.
How to use this Year 11 equation of a circle and tangents revision notes
First, read the key points so that you know where the Year 11 equation of a circle and tangents revision notes is going. Then work through each definition and worked example, and cover the solution before you read it. After that, practise with the equation of a circle and tangents worksheet and check every question with the worksheet answers. When you feel ready, test yourself with the multiple choice questions or the quick quiz. Finally, sit the equation of a circle and tangents test and mark it with the test answers.
Also, the starters are a good warm-up, so you can keep the key facts fresh. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, this Year 11 equation of a circle and tangents revision notes gives you everything you need for the chapter.
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