Circle

Circle

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Standard Equation: $(x-h)^2 + (y-k)^2 = r^2$, centre $(h,k)$, radius $r$.

General Form: $x^2 + y^2 + 2gx + 2fy + c = 0$. Centre: $(-g, -f)$. Radius: $\sqrt{g^2 + f^2 - c}$ (real if $g^2 + f^2 > c$).

Special Cases:

• Circle through origin: $c = 0$
• Circle with centre at origin: $x^2 + y^2 = r^2$
• Touches x-axis: centre $(h, r)$ or $(h, -r)$, so $(x-h)^2 + (y \mp r)^2 = r^2$

Parametric Form: $x = h + r\cos\theta, y = k + r\sin\theta$, where $\theta \in [0, 2\pi)$.

⚡ JEE Tip: For tangent to circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$, the equation is $xx_1 + yy_1 = r^2$. This is different from the chord formula.

⚡ Common Mistake: The radius formula $\sqrt{g^2+f^2-c}$ requires $g^2+f^2 > c$ for real radius. If $g^2+f^2 = c$, the circle is a point (radius 0). If $g^2+f^2 < c$, the circle is imaginary.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding.

Tangent Equations:

For circle $x^2 + y^2 = r^2$:

• At point $(x_1, y_1)$ on circle: $xx_1 + yy_1 = r^2$
• Slope form: $y = mx \pm r\sqrt{1+m^2}$
• Parametric: $x\cos\theta + y\sin\theta = r$

For circle $(x-h)^2 + (y-k)^2 = r^2$:

• At point $(x_1, y_1)$: $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$
• Slope form: more complex, derive from parametric

Normal Equations:

The normal at $(x_1, y_1)$ passes through centre $(h,k)$ and point $(x_1, y_1)$. So it’s the line through $(h,k)$ and $(x_1, y_1)$.

Chord of Contact: From external point $(x_1, y_1)$ to circle $x^2 + y^2 = r^2$: $xx_1 + yy_1 = r^2$.

Chord Bisected at a Point: For chord bisected at $(x_1, y_1)$: $T = S_1$: $xx_1 + yy_1 - r^2 = x_1^2 + y_1^2 - r^2$ → $xx_1 + yy_1 = x_1^2 + y_1^2$.

Worked Examples:

Example 1: Find the equation of the circle passing through $(1,1), (2,3), (3,2)$.

Let general circle: $x^2 + y^2 + 2gx + 2fy + c = 0$. At $(1,1)$: $1 + 1 + 2g + 2f + c = 0$ → $2 + 2g + 2f + c = 0$. At $(2,3)$: $4 + 9 + 4g + 6f + c = 0$ → $13 + 4g + 6f + c = 0$. At $(3,2)$: $9 + 4 + 6g + 4f + c = 0$ → $13 + 6g + 4f + c = 0$.

Subtract first from second: $11 + 2g + 4f = 0$ → $2g + 4f = -11$. Subtract first from third: $11 + 4g + 2f = 0$ → $4g + 2f = -11$. Solve: from first eq $g = -11/2 - 2f$. Plug into second: $4(-11/2 - 2f) + 2f = -11$ → $-22 - 8f + 2f = -11$ → $-6f = 11$ → $f = -11/6$. Then $g = -11/2 - 2(-11/6) = -11/2 + 11/3 = (-33 + 22)/6 = -11/6$. From first equation: $c = -2 - 2g - 2f = -2 - 2(-11/6) - 2(-11/6) = -2 + 11/3 + 11/3 = -2 + 22/3 = (-6+22)/3 = 16/3$.

So circle: $x^2 + y^2 - \frac{11}{3}x - \frac{11}{3}y + \frac{16}{3} = 0$. Multiply by 3: $3x^2 + 3y^2 - 11x - 11y + 16 = 0$.

Example 2 (JEE 2022): Find length of tangent from $(5, 3)$ to $x^2 + y^2 - 4x - 6y + 3 = 0$.

Circle: $g = -2, f = -3, c = 3$. Radius $r = \sqrt{4 + 9 - 3} = \sqrt{10}$. Centre $C = (2, 3)$. Distance from $P(5,3)$ to $C$: $PC = \sqrt{(5-2)^2 + (3-3)^2} = 3$. Length of tangent $PT = \sqrt{PC^2 - r^2} = \sqrt{9 - 10} = \sqrt{-1}$ — imaginary. This means the point $(5,3)$ lies inside the circle, so no real tangents exist.

Verify by the power test: substitute $(5,3)$ into $(x-2)^2 + (y-3)^2 - r^2$. $(3)^2 + 0 - 10 = 9 - 10 = -1 < 0$, confirming the point is inside the circle.

Example 3: Find the image (inverse) of point $(3, 4)$ in the circle $x^2 + y^2 = 25$.

The image of a point in a circle is its inverse: the point $Q$ on ray $OP$ such that $OP \cdot OQ = r^2$, where $O(0,0)$ is the centre. The line from $O(0,0)$ to $P(3,4)$ is $y = \frac{4}{3}x$. Here $OP = \sqrt{3^2 + 4^2} = 5$, and since $3^2 + 4^2 = 25$, the point $P$ lies on the circle. Applying the inversion relation: $OQ = \dfrac{r^2}{OP} = \dfrac{25}{5} = 5$. Since $Q$ lies on ray $OP$ at the same distance $5$ as $P$, we get $Q = P = (3,4)$. This illustrates a general property: every point on the circle of inversion is its own image.

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for serious JEE Advanced preparation.

Radical Axis and Radical Centre:

For two circles $S_1 = 0$ and $S_2 = 0$: Radical axis: $S_1 - S_2 = 0$ (linear equation, a line perpendicular to line of centres).

The radical axis is the locus of points with equal power with respect to both circles. For three circles, the three radical axes concur at the radical centre.

Common Tangents:

For two circles with centres $C_1, C_2$ and radii $r_1, r_2$:

• Direct common tangents: Do not cross the line of centres. There are 2.
• Transverse (cross) common tangents: Cross the line of centres. There are 2 if circles don’t overlap.
• Number of common tangents depends on distance $d = C_1C_2$:
  • $d > r_1 + r_2$: 4 tangents (external don’t intersect segment $C_1C_2$)
  • $d = r_1 + r_2$: 3 tangents (external tangent touches both)
  • $r_1 + r_2 > d > |r_1 - r_2|$: 2 tangents
  • $d = |r_1 - r_2|$: 1 tangent (internal)
  • $d < |r_1 - r_2|$: 0 tangents (one inside other)

Advanced Problems:

JEE Advanced 2020: Find the number of common tangents to circles $x^2 + y^2 = 1$ and $(x-4)^2 + y^2 = 9$.

Circle 1: centre $O(0,0), r_1 = 1$. Circle 2: centre $C(4,0), r_2 = 3$. $d = OC = 4$. $r_1 + r_2 = 1 + 3 = 4$. So $d = r_1 + r_2$. Therefore exactly 3 common tangents.

JEE Advanced 2019: Find the equation of the circle which touches the circle $x^2 + y^2 = 4$ externally at $(2,0)$ and has radius 3.

Let the required circle have centre $C_1(h,k)$ and radius $r_1 = 3$. It touches the circle $S: x^2+y^2=4$ externally at $P(2,0)$, so $P$ lies on both circles and the contact point lies on the line of centres.

For two circles touching externally at $P$, the centres lie on opposite sides of $P$ along the line of centres, with $OP = r_O$, $C_1P = r_1$, and $OC_1 = r_O + r_1$. Here circle $S$ has centre $O(0,0)$ and radius $r_O = 2$, so $OP = 2$, $C_1P = 3$, and $OC_1 = 2 + 3 = 5$.

The contact point $P=(2,0)$ lies on the positive x-axis from $O$. The centre $C_1$ lies on this same ray, beyond $P$, at distance $5$ from $O$. Thus $C_1 = (5, 0)$, and indeed the distance from $C_1$ to $P$ is $5 - 2 = 3 = r_1$. ✓

So the required circle is: centre $(5,0)$, radius $3$: $(x-5)^2 + y^2 = 9$.

JEE Advanced Patterns (2018–2024):

• Radical axis problems and properties of concurrent radical axes appeared in 2020, 2023
• Number of common tangents between circles is frequently tested
• Family of circles (touching conditions) is a major topic
• Orthogonal circles (where tangents are perpendicular at intersection) appeared in 2021
• Combined problems with lines and circles are very common

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