3. Introduction (definition)
Circle: The set of all points
that are the same distance
(equidistant) from a fixed
point.
Center: the fixed points
Radius: a segment whose
endpoints are the center
and a point on the circle
4. You can write an
equation of a circle in a
coordinate plane if you
know its radius and the
coordinates of its
center.
5. Suppose the
radius is r and
the center is (h,
k). Let (x, y) be
any point on the
circle. The
distance between
(x, y) and (h, k) is
r, so you can use
the Distance
Formula.
6. 1. The equation of circle centered at (0,0) and with radius r
9. Proof How could we show how
, and are related?
(Hint: draw a right-angled triangle
inside your circle, with one vertex at
the origin and another at the
circumference)
10. 2. The equation of circle centered at (a,b) and with radius r
Now suppose we shift
the circle so it’s now
centred at (a,b). What’s
the equation now?
(Hint: What would the sides of this
right-angled triangle be now?)
18. Equations of Circle (Key Concept)
1) The standard form: (x - a)2
+ (y-b)2
= r2
2) The general form : x^2 + y^2 + Ax + By + C = 0, A ≠ 0, and B ≠ 0.
The centre of the circle is and the radius is
If any one of the following three sets of facts about a circle is known,
then the other two can be determined:
center (h, k) and radius r
19.
20. Definition of a Tangent
A line or line segment touching the circle at
one point.
26. As always, to find an
equation we need (i) a point
and (ii) the gradient.
Point: (4,2)
Gradient?
Gradient of radius is 𝟐/𝟒=𝟏/𝟐
Gradient of tangent =−𝟐
𝒚−𝟐=−𝟐(𝒙−𝟒)
𝒚=−𝟐𝒙+𝟏𝟎
27. Gradient of chord: −𝟒/𝟐=−𝟐
Midpoint of chord: (1,2)
Gradient of radius =𝟏/𝟐
Equation of radius:
𝒚−𝟐=𝟏/𝟐 (𝒙−𝟏)
If 𝒙=𝟔:
𝒚−𝟐=𝟏/𝟐 (𝟔−𝟏)
𝒚=𝟒.𝟓
28. Question 1: The sketch show point 𝑃
on a circle, centre C. The equation of
the tangent at P is 𝑦=13−2𝑥.
Work out the gradient of PC. 𝟏/𝟐
Work out the equation of the circle.
Equation of PC:
𝒚−𝟓=𝟏/𝟐 (𝒙−𝟒)
𝑪(𝟎,𝟑) → 𝒙^𝟐+(𝒚−𝟑)^𝟐=𝟐𝟎
31. Question 4: The sketch shows part of a circle, centre C, that
intersects the axes at points and .
(a)Explain why the centre of the circle lies on the line
Because is the midpoint of and the perpendicular
bisector of QR goes through the centre of the circle.
(b)Show that the line is the perpendicular bisector of the
line .
Midpoint of : (2,2). Gradient: . Therefore perp bis.:
(c)Work out the equation of the circle.
Centre: (7,7) Radius:
32. Question 1: The points and form a diameter of a circle. The point is
another point on the circle.
(a) Determine the gradient of the line .
(b) Hence determine the equation of the line .
(c) Hence, given that the point lies on this line, determine the value of .
Question 2: A line with equation is tangent to a circle at the point .
(a)Determine the equation of .
(b)Determine the value of .
(c) Hence determine the equation of the circle.