Co-ordinate Geometry The Circle

Leaving Cert Higher Level Mathematics revision notes with diagrams, key terms and self-check questions.

6 min readHigher LevelBy Studytok
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Coordinate geometry of the circle bridges algebra and geometry by turning circular paths into equations on the Cartesian plane. In Leaving Certificate Higher Level Mathematics, Paper 2 routinely tests your ability to move between standard and general circle forms, determine whether lines intersect or touch a circle, construct tangent equations, and analyse circles that touch each other or the coordinate axes.

Standard Form and General Form

A circle is defined as the set of all points (x,y)(x, y) that sit at a fixed radius length rr from a fixed centre point. When the centre is the origin (0,0)(0, 0), applying Pythagoras' theorem gives x2+y2=r2x^2 + y^2 = r^2. Learn this equation and understand where it comes from (the distance from the centre is always r). Check your own copy of the Formulae and Tables booklet to see which circle formulas it lists.

When the centre shifts to (h,k)(h, k), the distance formula gives the standard form (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. Notice how the signs flip inside the brackets: an equation reading (x5)2+(y+3)2=49(x - 5)^2 + (y + 3)^2 = 49 has centre (5,3)(5, -3) and radius r=7r = 7.

A circle centred at C(h,k), with radius to P(x,y) forming the hypotenuse of a right triangle.
A circle centred at C(h,k), with radius to P(x,y) forming the hypotenuse of a right triangle.

Expanding (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2 gives x2+y22hx2ky+(h2+k2r2)=0x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0. This expansion is usually written in the general form:

x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0

Comparing coefficients shows that 2g=2h2g = -2h and 2f=2k2f = -2k, so the centre is (g,f)(-g, -f). The constant term is c=g2+f2r2c = g^2 + f^2 - r^2, which rearranges to give the radius r=g2+f2cr = \sqrt{g^2 + f^2 - c}. Before extracting gg, ff, or cc, always divide across by any common coefficient attached to x2x^2 and y2y^2 so that both squared terms start with 11.

Testing Points and Axis Intercepts

To determine whether a point (x1,y1)(x_1, y_1) sits inside, outside, or on a circle, you compare its distance from the centre with the radius. When working with the standard form, substitute the point into (x1h)2+(y1k)2(x_1 - h)^2 + (y_1 - k)^2. If the result is less than r2r^2, the point is inside; if it equals r2r^2, it lies on the boundary; if it exceeds r2r^2, it is outside.

If the circle is given in general form, substitute the point directly into the expression x12+y12+2gx1+2fy1+cx_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c. A negative value means the point is inside, zero means on the circle, and a positive value means outside. For example, testing (5,4)(5, 4) in x2+y2+2x6y15=0x^2 + y^2 + 2x - 6y - 15 = 0 gives 25+16+102415=12>025 + 16 + 10 - 24 - 15 = 12 > 0, confirming that (5,4)(5, 4) is outside.

To find where a circle crosses the coordinate axes, set the opposing coordinate to zero. Letting y=0y = 0 leaves a quadratic in xx whose solutions are the xx-intercepts. Similarly, setting x=0x = 0 yields the yy-intercepts. When a circle touches the xx-axis, the radius equals f|-f|, so g2+f2c=f2g^2 + f^2 - c = f^2, which simplifies to the useful condition g2=cg^2 = c.

A circle below the x-axis touches it at T(−g,0); its vertical radius from C(−g,−f) has length |−f|.
A circle below the x-axis touches it at T(−g,0); its vertical radius from C(−g,−f) has length |−f|.

Lines and Circles: Intersections and Tangency

A line and a circle can cross at two distinct points, touch at one single point of contact as a tangent, or miss each other entirely. You can find exact intersection points by rearranging the linear equation for one variable, substituting that expression into the circle equation, solving the resulting quadratic equation in one variable, and substituting each solution back into the line equation to find the matching co-ordinate.

For example, to find where the line xy+1=0x - y + 1 = 0 meets the circle x2+y2=25x^2 + y^2 = 25:

  1. From the line, express yy in terms of xx: y=x+1y = x + 1.
  2. Substitute into the circle equation: x2+(x+1)2=25x^2 + (x + 1)^2 = 25, which gives 2x2+2x24=02x^2 + 2x - 24 = 0, or x2+x12=0x^2 + x - 12 = 0.
  3. Factorise: (x+4)(x3)=0(x + 4)(x - 3) = 0, so x=3x = 3 or x=4x = -4.
  4. Substitute back into y=x+1y = x + 1: x=3y=4x = 3 \Rightarrow y = 4, and x=4y=3x = -4 \Rightarrow y = -3. The points of intersection are (3,4)(3, 4) and (4,3)(-4, -3) (check: 32+42=253^2 + 4^2 = 25).

The quadratic discriminant b24acb^2 - 4ac tells you the geometric arrangement without needing to find the coordinates. If b24ac>0b^2 - 4ac > 0, the line crosses the circle twice. If b24ac=0b^2 - 4ac = 0, the line is a tangent touching at one point. If b24ac<0b^2 - 4ac < 0, the line does not meet the circle.

When a problem asks you to prove tangency or find an unknown constant, using the perpendicular distance formula is almost always faster than algebra substitution. A straight line is tangent to a circle if and only if the perpendicular distance dd from the centre (x1,y1)(x_1, y_1) to the line ax+by+c=0ax + by + c = 0 equals the radius rr:

d=ax1+by1+ca2+b2=rd = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} = r
Three panels show a secant, tangent and non-intersecting line, with perpendicular distances smaller than, equal to and greater than the radius.
Three panels show a secant, tangent and non-intersecting line, with perpendicular distances smaller than, equal to and greater than the radius.

Constructing Tangent Equations

Finding a tangent depends on whether the given point lies on the circle or outside it. By geometric theorem, every tangent line is perpendicular to the radius drawn to the point of contact.

If the point of contact (x1,y1)(x_1, y_1) is on the circle, first find the slope of the radius connecting the centre to that point using mr=y1kx1hm_r = \frac{y_1 - k}{x_1 - h}. The tangent slope is its negative reciprocal, mt=1mrm_t = -\frac{1}{m_r}. Then write the line equation using yy1=mt(xx1)y - y_1 = m_t(x - x_1).

If the point (x1,y1)(x_1, y_1) lies outside the circle, write the tangent as an unknown line through that point: yy1=m(xx1)y - y_1 = m(x - x_1), which rearranges to mxy+(y1mx1)=0mx - y + (y_1 - mx_1) = 0. Apply the perpendicular distance formula from the centre to this line and set it equal to rr. Squaring both sides produces a quadratic equation in mm, yielding the two tangent slopes that pass through that external point.

Touching Circles and Tangent Lengths

Two circles touch each other when their boundaries meet at exactly one point. You test this by calculating the distance d=c1c2d = |c_1c_2| between their centres and comparing it with their radii r1r_1 and r2r_2.

For external contact, the two circles sit outside each other and touch at a single point between their centres. This occurs when d=r1+r2d = r_1 + r_2. For internal contact, one circle sits inside the other, which occurs when d=r1r2d = |r_1 - r_2|. In an exam, state both values explicitly and finish with a concluding sentence showing that they match.

Externally touching circles have the contact point between their centres; internally touching circles have it beyond both centres.
Externally touching circles have the contact point between their centres; internally touching circles have it beyond both centres.

Another classic exam question asks for the length of a tangent drawn from an external point PP to a circle with centre CC and radius rr. Because the radius to the contact point TT meets the tangent at 9090^\circ, triangle PTCPTC is right-angled at TT. Pythagoras' theorem gives PT2+r2=CP2|PT|^2 + r^2 = |CP|^2, so the tangent length is PT=CP2r2|PT| = \sqrt{|CP|^2 - r^2}.

Two tangents from external point P touch a circle at T and U. Triangle PTC is right-angled at T, with CP as its hypotenuse.
Two tangents from external point P touch a circle at T and U. Triangle PTC is right-angled at T, with CP as its hypotenuse.

Key terms

Standard Form
The circle equation (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the centre and rr is the radius.
General Form
The circle equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, with centre (g,f)(-g, -f) and radius r=g2+f2cr = \sqrt{g^2 + f^2 - c}.
Point of Contact
The single point where a tangent line touches a circle, forming a right angle with the radius drawn to that point.
Perpendicular Distance
The shortest distance from a point to a straight line; a line is tangent to a circle if and only if this distance from the centre equals the radius.
External Contact
The geometric condition where two separate circles touch at one boundary point, satisfying c1c2=r1+r2|c_1c_2| = r_1 + r_2.
Internal Contact
The geometric condition where one circle sits inside another and touches it at one point, satisfying c1c2=r1r2|c_1c_2| = |r_1 - r_2|.

Check yourself

  1. What are the centre and radius of the circle 2x2+2y212x+16y=02x^2 + 2y^2 - 12x + 16y = 0?

    Divide by 2 to get x2+y26x+8y=0x^2 + y^2 - 6x + 8y = 0. Here g=3g = -3, f=4f = 4, and c=0c = 0. The centre is (g,f)=(3,4)(-g, -f) = (3, -4) and the radius is (3)2+420=25=5\sqrt{(-3)^2 + 4^2 - 0} = \sqrt{25} = 5.

  2. Determine whether the point (2,3)(2, 3) lies inside, outside, or on the circle x2+y2=13x^2 + y^2 = 13.

    Substitute the point into x2+y2x^2 + y^2: 22+32=4+9=132^2 + 3^2 = 4 + 9 = 13. Because the value equals r2=13r^2 = 13, the point lies exactly on the circle.

  3. Two circles have radii of 8 and 3, and their centres are 5 units apart. Do they touch externally or internally?

    They touch internally because the distance between their centres equals the difference between their radii: 83=5|8 - 3| = 5.

  4. What algebraic condition links gg and cc when the circle x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0 touches the xx-axis?

    The radius equals the distance to the xx-axis, meaning r=fr = |-f|. Since r2=g2+f2cr^2 = g^2 + f^2 - c, setting this equal to f2f^2 gives g2=cg^2 = c.

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