Circle – Equation, Properties, Tangents, and Applications

TAKEAWAYS:

• The standard equation of a circle with center $(h,k)$ and radius $r$ is $(x-h{)}^2+(y-k{)}^2=r^2$.
• A tangent to a circle touches it at exactly one point and is perpendicular to the radius at that point.
• The length of a tangent from an external point $(x_1,y_1)$ to a circle $(x-h{)}^2+(y-k{)}^2=r^2$ is $\sqrt{(x_1-h{)}^2+(y_1-k{)}^2-r^2}$.
• The equation of a tangent to a circle can be derived using the condition of tangency (discriminant = 0).
• Circles can intersect, touch, or be disjoint; their relative positions are determined by the distance between centers and radii.
• Applications include designing wheels, lenses, and satellite orbits.

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### **1. Definition and Standard Equation of a Circle**
A **circle** is the set of all points in a plane that are at a fixed distance (radius) from a fixed point (center).

```figure
{"type":"circle","center":"(h, k)","radius":"r","chord":true,"angleAtCentre":"2θ","angleAtCircumference":"θ","tangent":true,"points":[{"x":"h + r","y":"k","label":"P"},{"x":"h","y":"k + r","label":"Q"}],"caption":"Circle with center (h, k), radius r, showing central angle 2θ, inscribed angle θ, and tangent at P"}

Standard Equation

The standard equation of a circle with center ((h, k)) and radius (r) is: [ (x - h)^2 + (y - k)^2 = r^2 ]

• If the circle is centered at the origin ((0, 0)), the equation simplifies to: [ x^2 + y^2 = r^2 ]

Derivation

Consider a circle with center ((h, k)) and radius (r). Let ((x, y)) be any point on the circle. The distance between ((x, y)) and ((h, k)) is (r): [ \sqrt{(x - h)^2 + (y - k)^2} = r ] Squaring both sides gives the standard equation.

Worked Example 1: Write the equation of a circle with center ((-2, 3)) and radius (5).

Solution: Using the standard form: [ (x - (-2))^2 + (y - 3)^2 = 5^2 ] [ (x + 2)^2 + (y - 3)^2 = 25 ]

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2. General Equation of a Circle

The general equation of a circle is: [ x^2 + y^2 + 2gx + 2fy + c = 0 ] where:

• Center = ((-g, -f))
• Radius (r = \sqrt{g^2 + f^2 - c})

Worked Example 2: Find the center and radius of the circle (x^2 + y^2 - 4x + 6y + 4 = 0).

Solution: Compare with the general form: [ 2g = -4 \implies g = -2 ] [ 2f = 6 \implies f = 3 ] [ c = 4 ] Center = ((-g, -f) = (2, -3)) Radius: [ r = \sqrt{(-2)^2 + 3^2 - 4} = \sqrt{4 + 9 - 4} = \sqrt{9} = 3 ]

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3. Position of a Point Relative to a Circle

For a circle ((x - h)^2 + (y - k)^2 = r^2) and a point ((x_1, y_1)), the position is determined by: [ d = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} ]

• If (d < r), the point is inside the circle.
• If (d = r), the point is on the circle.
• If (d > r), the point is outside the circle.

Worked Example 3: Determine where the point ((5, 1)) lies relative to the circle ((x - 1)^2 + (y + 2)^2 = 25).

Solution: Center ((h, k) = (1, -2)), radius (r = 5). Distance: [ d = \sqrt{(5 - 1)^2 + (1 - (-2))^2} = \sqrt{16 + 9} = 5 ] Since (d = r), the point lies on the circle.

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4. Tangents to a Circle

A tangent to a circle is a line that touches the circle at exactly one point. It is perpendicular to the radius at the point of contact.

Equation of Tangent

For a circle ((x - h)^2 + (y - k)^2 = r^2), the tangent at a point ((x_1, y_1)) on the circle is: [ (x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2 ]

Worked Example 4: Find the equation of the tangent to the circle (x^2 + y^2 = 25) at the point ((3, 4)).

Solution: Here, (h = 0), (k = 0), (r = 5), and ((x_1, y_1) = (3, 4)). Using the tangent formula: [ (3 - 0)(x - 0) + (4 - 0)(y - 0) = 25 ] [ 3x + 4y = 25 ]

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5. Length of Tangent from an External Point

The length of the tangent from an external point ((x_1, y_1)) to the circle ((x - h)^2 + (y - k)^2 = r^2) is: [ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} ]

Worked Example 5: Find the length of the tangent from ((8, 6)) to the circle ((x - 2)^2 + (y - 4)^2 = 25).

Solution: Here, ((h, k) = (2, 4)), (r = 5), and ((x_1, y_1) = (8, 6)). [ L = \sqrt{(8 - 2)^2 + (6 - 4)^2 - 25} = \sqrt{36 + 4 - 25} = \sqrt{15} ]

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6. Condition for Tangency

A line (y = mx + c) is tangent to the circle (x^2 + y^2 + 2gx + 2fy + c = 0) if the discriminant of the combined equation is zero.

Worked Example 6: Show that the line (y = 2x + 3) is tangent to the circle (x^2 + y^2 - 4x - 6y + 9 = 0).

Solution: Substitute (y = 2x + 3) into the circle’s equation: [ x^2 + (2x + 3)^2 - 4x - 6(2x + 3) + 9 = 0 ] [ x^2 + 4x^2 + 12x + 9 - 4x - 12x - 18 + 9 = 0 ] [ 5x^2 - 4x = 0 ] Discriminant (D = (-4)^2 - 4(5)(0) = 16 > 0)? Wait, this seems incorrect! Correction: The correct substitution should yield a quadratic in (x) with (D = 0) for tangency. Let’s re-solve properly.

Substitute (y = 2x + 3) into (x^2 + y^2 - 4x - 6y + 9 = 0): [ x^2 + (2x + 3)^2 - 4x - 6(2x + 3) + 9 = 0 ] [ x^2 + 4x^2 + 12x + 9 - 4x - 12x - 18 + 9 = 0 ] [ 5x^2 - 4x = 0 ] This gives (x(5x - 4) = 0), so two solutions ((x = 0) and (x = 4/5)), meaning the line intersects the circle at two points. This contradicts the problem statement!

Re-evaluating the circle’s equation: The given circle is (x^2 + y^2 - 4x - 6y + 9 = 0). Let’s rewrite it in standard form: [ x^2 - 4x + y^2 - 6y = -9 ] Complete the square: [ (x^2 - 4x + 4) + (y^2 - 6y + 9) = -9 + 4 + 9 ] [ (x - 2)^2 + (y - 3)^2 = 4 ] Now, the circle has center ((2, 3)) and radius (2).

Distance from center to line (y = 2x + 3) (or (2x - y + 3 = 0)): [ d = \frac{|2(2) - 1(3) + 3|}{\sqrt{2^2 + (-1)^2}} = \frac{|4 - 3 + 3|}{\sqrt{5}} = \frac{4}{\sqrt{5}} \approx 1.789 ] Since (d > r) (radius = 2), the line does not intersect the circle. This suggests the original problem may have an error.

Alternative Approach: Assume the circle is (x^2 + y^2 - 4x - 6y + 12 = 0) (radius = 3). Rewriting: [ (x - 2)^2 + (y - 3)^2 = 9 ] Distance from ((2, 3)) to (2x - y + 3 = 0): [ d = \frac{|4 - 3 + 3|}{\sqrt{5}} = \frac{4}{\sqrt{5}} \approx 1.789 < 3 ] Still not tangent. Conclusion: The line (y = 2x + 3) is not tangent to the given circle. Verify the problem statement.

Correct Example: Find the condition for the line (y = mx + c) to be tangent to (x^2 + y^2 = r^2). Substitute (y = mx + c): [ x^2 + (mx + c)^2 = r^2 ] [ (1 + m^2)x^2 + 2mcx + (c^2 - r^2) = 0 ] For tangency, discriminant (D = 0): [ (2mc)^2 - 4(1 + m^2)(c^2 - r^2) = 0 ] [ 4m^2c^2 - 4(c^2 - r^2 + m^2c^2 - m^2r^2) = 0 ] [ c^2 = r^2(1 + m^2) ]

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7. Chord of Contact

The chord of contact is the line joining the points of contact of the two tangents drawn from an external point ((x_1, y_1)) to the circle ((x - h)^2 + (y - k)^2 = r^2). Its equation is: [ (x_1 - h)(x - h) + (y_1 - k)(y - k) = r^2 ]

Worked Example 7: Find the chord of contact from ((4, 2)) to the circle (x^2 + y^2 = 16).

Solution: Here, (h = 0), (k = 0), (r = 4), and ((x_1, y_1) = (4, 2)). Chord of contact: [ (4 - 0)(x - 0) + (2 - 0)(y - 0) = 16 ] [ 4x + 2y = 16 \implies 2x + y = 8 ]

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8. Director Circle

The director circle of a given circle is the locus of points from which tangents to the circle are perpendicular to each other. For a circle ((x - h)^2 + (y - k)^2 = r^2), its director circle is: [ (x - h)^2 + (y - k)^2 = 2r^2 ]

Worked Example 8: Find the director circle of (x^2 + y^2 = 25).

Solution: Here, (h = 0), (k = 0), (r = 5). Director circle: [ x^2 + y^2 = 2(5)^2 = 50 ]

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9. Family of Circles

The family of circles passing through the intersection of two circles (C_1 = 0) and (C_2 = 0) is given by: [ C_1 + \lambda C_2 = 0 ] where (\lambda) is a parameter.

Worked Example 9: Find the equation of the family of circles passing through the intersection of (x^2 + y^2 - 4x = 0) and (x^2 + y^2 - 6x - 8y = 0).

Solution: Let (C_1 = x^2 + y^2 - 4x = 0) and (C_2 = x^2 + y^2 - 6x - 8y = 0). Family of circles: [ C_1 + \lambda C_2 = 0 \implies (1 + \lambda)x^2 + (1 + \lambda)y^2 - (4 + 6\lambda)x - 8\lambda y = 0 ] Divide by ((1 + \lambda)): [ x^2 + y^2 - \left(\frac{4 + 6\lambda}{1 + \lambda}\right)x - \left(\frac{8\lambda}{1 + \lambda}\right)y = 0 ]

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10. Orthogonal Circles

Two circles are orthogonal if the angle between their tangents at the points of intersection is (90^\circ). The condition for orthogonality is: [ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 ] where the circles are: [ C_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 ] [ C_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 ]

Worked Example 10: Check if the circles $x^2+y^2-4x-6y+4=0$ and $x^2+y^2-6x-4y+4=0$ are orthogonal.

Solution: For $C_1$: $2g_1=-4⟹g_1=-2$, $2f_1=-6⟹f_1=-3$, $c_1=4$. For $C_2$: $2g_2=-6⟹g_2=-3$, $2f_2=-4⟹f_2=-2$, $c_2=4$. Check orthogonality: $$2(-2)(-3)+2(-3)(-2)=12+12=24$$ $$c_1+c_2=4+4=8$$ Since $24\ne 8$, the circles are not orthogonal.

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11. Applications of Circles

1. Engineering: Designing wheels, gears, and pipes.
2. Physics: Orbital paths of planets and satellites.
3. Computer Graphics: Drawing circles and arcs.
4. Architecture: Dome structures and circular arches.

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Exam Tip

1. Memorize the standard and general equations of a circle.
2. Practice deriving the equation of a tangent and chord of contact.
3. Use the condition for tangency (discriminant = 0) to solve problems.
4. Understand the geometric interpretations (e.g., director circle, orthogonal circles).
5. Solve numerical problems involving distances and lengths related to circles.
6. Draw diagrams for every problem to visualize the scenario.

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NEB Board-Style Questions

Short Answer Questions

1. Write the equation of a circle with center $(-1,2)$ and radius $3$.
2. Find the center and radius of the circle $x^2+y^2+6x-8y+21=0$.
3. Determine if the point $(4,-1)$ lies inside, on, or outside the circle $(x+2{)}^2+(y-3{)}^2=25$.
4. Find the equation of the tangent to the circle $x^2+y^2=10$ at the point $(1,3)$.
5. Calculate the length of the tangent from $(5,6)$ to the circle $(x-2{)}^2+(y+1{)}^2=16$.

Long Answer Questions

6. Prove that the line $y=mx+c$ is tangent to the circle $x^2+y^2=r^2$ if $c^2=r^2(1+m^2)$.
7. Find the equation of the chord of contact from the point $(3,4)$ to the circle $x^2+y^2-6x-8y+21=0$.
8. Show that the circles $x^2+y^2-4x-6y+9=0$ and $x^2+y^2-6x-4y+4=0$ intersect orthogonally.
9. Find the equation of the family of circles passing through the intersection of $x^2+y^2-x=0$ and $x^2+y^2-y=0$.
10. A circle has center $(2,-3)$ and passes through $(4,1)$. Find its equation and the length of the tangent from $(0,0)$ to this circle.

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Note: Always draw diagrams for circle problems to visualize centers, radii, tangents, and points. Practice converting between standard and general forms, and use the condition for tangency in problems involving lines and circles.