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Grade 12 · Analytical Geometry · 11 min read

The Circle & Its Tangent (Grade 12)

The equation (x − a)² + (y − b)² = r², completing the square to find the centre and radius from the general form, and finding the tangent at a point on the circle.

1The standard form

(x − a)² + (y − b)² = r²

centre (a ; b), radius r
⚠️

The centre is (a ; b), and the signs flip. (x − 2)² + (y + 3)² = 25 has centre (2 ; −3), not (2 ; 3).

M(2 ; -3)P(6 ; 0)xy
x² + y² − 4x + 6y − 12 = 0: centre M(2 ; −3), radius 5, with P(6 ; 0) on the circle.

2From the general form: complete the square

Worked ExampleDetermine the centre and radius of x² + y² − 4x + 6y − 12 = 0
−4−22468−8−6−4−22M(2 ; −3)xycomplete the square to find M and r
  1. 1
    Group: (x² − 4x) + (y² + 6y) = 12.
    Move the constant to the right.
  2. 2
    Halve and square each: (−4 ÷ 2)² = 4 and (6 ÷ 2)² = 9.
  3. 3
    (x² − 4x + 4) + (y² + 6y + 9) = 12 + 4 + 9.
    Add to BOTH sides.
  4. 4
    (x − 2)² + (y + 3)² = 25.
  5. 5
    Centre (2 ; −3), radius 5.
    r = 25 = 5.

3The tangent at a point

tangent ⊥ radius  ⇒  mtangent × mradius = −1
Worked ExampleDetermine the equation of the tangent to the circle above at P(6 ; 0)
M(2 ; -3)P(6 ; 0)xy
  1. 1
    Check P is on the circle: (6−2)² + (0+3)² = 16 + 9 = 25. ✓
    Always verify before you start.
  2. 2
    mMP = 0 − (−3)6 − 2 = 34.
    Gradient of the radius from the centre to P.
  3. 3
    mtangent = −43.
    Negative reciprocal, because the tangent is perpendicular to the radius.
  4. 4
    y − 0 = −43(x − 6).
    Point-gradient form through P.
  5. 5
    y = −43x + 8.
💡

Every circle question that mentions a tangent is really asking about perpendicular gradients. Find the radius gradient first, then flip and change the sign.

Practice exercises

Work each one out, then click to reveal the answer.

  1. 1
    Write down the centre and radius of (x − 2)² + (y + 3)² = 25.
    −4−22468−8−6−4−22Mxy
    Show answer ▾
    Centre (2 ; −3), radius 5
  2. 2
    Determine the centre and radius of x² + y² − 4x + 6y − 12 = 0.
    M(2 ; -3)P(6 ; 0)xy
    Show answer ▾
    Completing the square: (x−2)² + (y+3)² = 25 → centre (2 ; −3), r = 5
  3. 3
    Show that P(6 ; 0) lies on the circle x² + y² − 4x + 6y − 12 = 0.
    M(2 ; -3)P(6 ; 0)xy
    Show answer ▾
    (6−2)² + (0+3)² = 16 + 9 = 25 = r², so P lies on the circle.
  4. 4
    Determine the gradient of the radius MP where M(2 ; −3) and P(6 ; 0).
    −4−22468−8−6−4−22M(2 ; −3)P(6 ; 0)xy
    Show answer ▾
    34
  5. 5
    Determine the equation of the tangent to the circle at P(6 ; 0).
    M(2 ; -3)P(6 ; 0)xy
    Show answer ▾
    m = −43y = −43x + 8
  6. 6
    Write down the equation of a circle with centre (0 ; 0) and radius 7.
    −8−6−4−22468−8−6−4−22468M(0 ; 0)xy
    Show answer ▾
    x² + y² = 49
  7. 7
    Determine the radius of x² + y² + 2x − 8y + 8 = 0.
    −5−4−3−2−1123−112345678Mxy
    Show answer ▾
    (x+1)² + (y−4)² = −8 + 1 + 16 = 9 → r = 3
  8. 8
    What is the relationship between a tangent and the radius at the point of contact?
    M(2 ; -3)P(6 ; 0)xy
    Show answer ▾
    They are perpendicular.
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