Kenneth ChengMathematical modeling
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Equations of lines and circles: mixed review

Read the idea, work independently, then explain what changed.

TOPIC 01

Equations of lines and circles: mixed review

A mixed assessment: identify the method, justify it and revise your reasoning.

What you will be able to explain

  • Connect the chapter skills without relying on the order of the exercises.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

Find the slope of AB.

A=(1,2), B=(3,42)A=(1,2),\ B=(3,42)
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
m=Δy/Δxm=\Delta y/\Delta x
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m=(42−2)/(3−1)=20m=(42-2)/(3-1)=20

  3. Use consistent order in numerator and denominator.

The requested value is 20.

Checks and common pitfalls: Use consistent order in numerator and denominator.

Think first. Reveal a hint when the class is ready.

02 / Foundation#Your turn

Find the line through (t,1) and (t,5).

t=21t=21
  • Point-slope form omits vertical lines; intercept form requires nonzero intercepts.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
x1=x2x_1=x_2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    x=21x=21

  3. Use a vertical equation directly instead of dividing by zero in a slope formula.

The requested relation or conclusion is shown below.

x=21x=21

Checks and common pitfalls: Use a vertical equation directly instead of dividing by zero in a slope formula.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

03 / Foundation#Your turn

Find the distance from P to the line.

P=(0,0),3x+4y−110=0P=(0,0),\quad 3x+4y-110=0
  • The denominator is the norm of the normal vector; parallel-line formulas require matched coefficients.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
d=∣C∣/5d=|C|/5
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    d=110/9+16=22d=110/\sqrt{9+16}=22

  3. The normal (3,4) has length 5.

The requested value is 22.

Checks and common pitfalls: The normal (3,4) has length 5.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Find the circle centered at (t,−1) and passing through (t+3,3).

t=23t=23
  • In x²+y²+Dx+Ey+F=0, a nondegenerate circle requires (D²+E²)/4−F>0.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
r2=32+42r^2=3^2+4^2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    r2=25r^2=25

  3. Use the center-to-point distance; the point itself is not the center.

The requested relation or conclusion is shown below.

(x−23)2+(y+1)2=25(x-23)^2+(y+1)^2=25

Checks and common pitfalls: Use the center-to-point distance; the point itself is not the center.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Find the number of common points.

x2+y2=576,y=24x^2+y^2=576,\quad y=24
  • Compare center distance with r1+r2 and |r1−r2|; equal centers require separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
d=rd=r
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    d=24=r⇒tangentd=24=r\Rightarrow\text{tangent}

  3. Equality gives one tangent point.

The requested value is 1.

Checks and common pitfalls: Equality gives one tangent point.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

A line has inclination 135°. Find its slope.

m=tan⁡135∘m=\tan135^\circ
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
tan⁡(180∘−45∘)=−tan⁡45∘\tan(180^\circ-45^\circ)=-\tan45^\circ
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    m=−1m=-1

  3. Inclination lies in [0°,180°); a second-quadrant angle has negative tangent.

The requested value is -1.

Checks and common pitfalls: Inclination lies in [0°,180°); a second-quadrant angle has negative tangent.

Think first. Reveal a hint when the class is ready.

07 / Standard#Your turn

Find C so that Ax+By+C=0 passes through P.

A=2, B=−3, P=(26,1)A=2,\ B=-3,\ P=(26,1)
  • Point-slope form omits vertical lines; intercept form requires nonzero intercepts.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
C=−AxP−ByPC=-Ax_P-By_P
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    C=−2(26)+3=−49C=-2(26)+3=-49

  3. A point satisfies the equation exactly when the substituted expression is zero.

The requested value is -49.

Checks and common pitfalls: A point satisfies the equation exactly when the substituted expression is zero.

Think first. Reveal a hint when the class is ready.

08 / Standard#Your turn

Find the foot of the perpendicular from P to the line.

P=(27,2),y=xP=(27,2),\quad y=x
  • The denominator is the norm of the normal vector; parallel-line formulas require matched coefficients.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
H=(h,h),(P−H)⋅(1,1)=0H=(h,h),\quad(P-H)\cdot(1,1)=0
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    27+2−2h=027+2-2h=0

  3. Apply the stated relation and retain its conditions.

    h=14.5h=14.5

  4. A foot lies on the line and its displacement from P is perpendicular to the line.

The requested relation or conclusion is shown below.

H=(14.5,14.5)H=(14.5,14.5)

Checks and common pitfalls: A foot lies on the line and its displacement from P is perpendicular to the line.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

09 / Standard#Your turn

Find the circle through O=(0,0), A=(2t,0), B=(0,2).

t=28t=28
  • In x²+y²+Dx+Ey+F=0, a nondegenerate circle requires (D²+E²)/4−F>0.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
F=0;D=−2t;E=−2F=0;\quad D=-2t;\quad E=-2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    3136+56D=0⇒D=−563136+56D=0\Rightarrow D=-56

  3. Apply the stated relation and retain its conditions.

    4+2E=0⇒E=−24+2E=0\Rightarrow E=-2

  4. Three noncollinear points determine one circle.

The requested relation or conclusion is shown below.

x2+y2−56x−2y=0x^2+y^2-56x-2y=0

Checks and common pitfalls: Three noncollinear points determine one circle.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

10 / Standard#Your turn

Classify the relation of these two circles.

C1:x2+y2=841,C2:(x−32)2+y2=9C_1:x^2+y^2=841, C_2:(x-32)^2+y^2=9
  • Compare center distance with r1+r2 and |r1−r2|; equal centers require separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
d=r1+r2d=r1+r2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    d=32=r1+r2d=32=r_1+r_2

  3. The circles are externally tangent.

The requested relation or conclusion is shown below.

d=r1+r2=32d=r_1+r_2=32

Checks and common pitfalls: The circles are externally tangent.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

11 / Standard#Your turn

Find a so that the two nonvertical lines are parallel.

y=(2a+1)x+3,y=61x−1y=(2a+1)x+3,\quad y=61x-1
  • A vertical line has no finite slope; m1m2=−1 applies only when both slopes exist.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
2a+1=2t+12a+1=2t+1
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    2a+1=61⇒a=302a+1=61\Rightarrow a=30

  3. Different intercepts ensure distinct parallel lines once slopes match.

The requested value is 30.

Checks and common pitfalls: Different intercepts ensure distinct parallel lines once slopes match.

Think first. Reveal a hint when the class is ready.

12 / Standard#Your turn

Find the reflection of the line in the y-axis.

y=31x+2y=31x+2
  • Point-slope form omits vertical lines; intercept form requires nonzero intercepts.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
x↦−xx\mapsto-x
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    y=31(−x)+2=−31x+2y=31(-x)+2=-31x+2

  3. Reflection changes the x-coordinate sign while leaving y unchanged.

The requested relation or conclusion is shown below.

y=−31x+2y=-31x+2

Checks and common pitfalls: Reflection changes the x-coordinate sign while leaving y unchanged.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

13 / Transfer#Your turn

Reflect P in the line y=x.

P=(32,2)P=(32,2)
  • The denominator is the norm of the normal vector; parallel-line formulas require matched coefficients.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
P+P′=2HP+P\prime=2H
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    H=(17,17)H=(17,17)

  3. Apply the stated relation and retain its conditions.

    P′=2H−P=(2,32)P'=2H-P=(2,32)

  4. Reflection across y=x swaps coordinates.

The requested relation or conclusion is shown below.

P′=(2,32)P'=(2,32)

Checks and common pitfalls: Reflection across y=x swaps coordinates.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

14 / Transfer#Your turn

The circle center is (t,0), radius 2. Find the largest x-coordinate on it.

t=33t=33
  • In x²+y²+Dx+Ey+F=0, a nondegenerate circle requires (D²+E²)/4−F>0.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
Use this intermediate relation.
∣x−t∣≤2|x-t|\le2
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    xmax⁡=33+2=35x_{\max}=33+2=35

  3. The extremal point lies on the horizontal diameter.

The requested value is 35.

Checks and common pitfalls: The extremal point lies on the horizontal diameter.

Think first. Reveal a hint when the class is ready.

15 / Transfer#Your turn

Find the tangent at the given point.

x2+y2=28900,P=(102,136)x^2+y^2=28900, P=(102,136)
  • Compare center distance with r1+r2 and |r1−r2|; equal centers require separate treatment.
Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1
Translate the geometric condition into a coordinate equation.
Hint 2
The radius is normal to the tangent.
Worked solution

  1. Translate the geometric condition into a coordinate equation.

  2. Apply the stated relation and retain its conditions.

    3(x−102)+4(y−136)=03(x-102)+4(y-136)=0

  3. Apply the stated relation and retain its conditions.

    3x+4y=8503x+4y=850

  4. The point is on the circle and fixes the tangent constant.

The requested relation or conclusion is shown below.

3x+4y=8503x+4y=850

Checks and common pitfalls: The point is on the circle and fixes the tangent constant.

Reasoning checklist · self / teacher assessment
  • State a valid definition or model and its assumptions.
  • Show the intermediate mathematical relations, not only the final claim.
  • Check exclusions, units or the interpretation of the result.

Think first. Reveal a hint when the class is ready.

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    Teacher preparation and assessment

    Question sequence

    • Ask students to name the relevant condition before calculating.

    Board plan

    • Compare valid methods and annotate their conditions.

    Anticipated thinking

    • A correct final value may still hide a missing assumption.

    Assessment checklist

    • Check the method, conditions, reasoning and interpretation separately.

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