SME
Grade 10 Math · Problem type
A point on the circle
Use this for “A circle has center (1, 2) and radius 5. Find x when the point (x, 5) is on it.”
Diagram
(x − 1)² + (y − 2)² = 25
Center (1, 2)
radius 5.
The point (5, 5) is on the curve.
This is an example. Type your own values to start a new problem.
(
x
−
h
)²
+
(
y
−
k
)²
=
r
2
x₂
Other point, x
worked out (not typed)
Formulas
(
x
−
h
)
2
+
(
y
−
k
)
2
=
r
2
(5 − 1)
2
+
(5 − 2)
2
=
5
2
x₂
=
2
×
h
−
x
−3 = 2 × 1 − 5
Assumptions
•
A point is on the circle when (x − h)² + (y − k)² = r².
•
Given y, solve for x: x = h ± √(r² − (y − k)²). Two points share each height, mirrored across x = h.
•
The radius is the distance from the center to any point on the circle.
Step-by-step
WE KNOW
h = 1, k = 2, r = 5, y = 5
FIND
point, x (x), other point, x (x₂)
Step 1 · Find point, x (x)
Use (x − h)² + (y − k)² = r²
Solve
the
circle’s
equation
for
x
:
two
points
at
that
height.
x
=
h
±
√(
r
2
−
(
y
−
k
)
2
)
x
=
1
±
√(
5
2
−
(5 − 2)
2
)
x
=
1
±
√
25 − 9
x
=
1
±
√
16
x
=
1
±
4
x
=
5
Step 2 · Find other point, x (x₂)
Use x₂ = 2 × h − x
The
other
point
is
the
mirror
image
across
x
=
h
.
x₂
=
2
×
1
−
5
x₂
=
−3
Check
Put the numbers back in. Both sides should match.
(5 − 1)
2
+
(5 − 2)
2
=
5
2
✓
−3 = 2 × 1 − 5 ✓
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Center and radius from the general form
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Arc length, sector area and radian measure
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