Use these Inter 2nd Year Maths 2B Formulas PDF Chapter 2 System of Circles to solve questions creatively.

## Intermediate 2nd Year Maths 2B System of Circles Formulas

**Definition:**

→ The angle between two intersecting circles is defined as the angle between the tangents at the point of intersection of the two circle’s. If θ is the angle between the circles, then

cos θ = \(\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\)

Here d = distance between the centres, r_{1}, r_{2} be their radii.

→ If θ is angle between the two circles.

x^{2} + y^{2} + 2gx + 2fy + c = 0 and x^{2} + y^{2} + 2g’x + 2f’y + c’ = 0, then

cos θ = \(\frac{-2 g g^{\prime}-2 f f^{\prime}+c+c^{\prime}}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

→ Circles cut orthogonally if

2g’g + 2f’f = c + c’ [∵ cos 90° = 0]

(or) d^{2} = r^{2}_{1} + r^{2}_{2} then also two circles cut orthogonally.

Theorem:

If d is the distance between the centers of two intersecting circles with radii r_{1}, r_{2} and θ is the angle between the circles then cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\).

Proof:

Let C_{1}, C_{2} be the centre s of the two circles S = 0, S’ = 0 with radii r_{1}, r_{2} respectively. Thus C_{1}C_{2} = d. Let P be a point of intersection of the two circles. Let PB, PA be the tangents of the circles S = 0, S’ = 0 respectively at P.

Now PC_{1} = r_{1}, PC_{2} = r_{2}, ∠APB = θ

Since PB is a tangent to the circle S = 0, ∠C_{1}PB = π/2

Since PA is a tangent to the circle S’ = 0, ∠C_{2}PA = π/2

Now ∠C_{1}PC_{2} = ∠C_{1}PB + ∠C_{2}PA – ∠APB = π/2 + π/2 – θ = π – θ

From ∆C_{1}PC_{2}, by cosine rule,

C_{1}^{2}C_{2}^{2} = PC_{1}^{2} + PC_{2}^{2} – 2PC_{1} . PC_{2} cos ∠C_{1}PC_{2} ⇒ d^{2} = r_{1}^{2} + r_{2}^{2} – 2r_{1}r_{2} cos(π – θ) ⇒ d^{2} = r_{1}^{2} + r_{2}^{2} + 2r_{1}r_{2} cos θ

⇒ 2r_{1}r_{2} cos θ = d^{2} – r_{1}^{2} – r_{2}^{2} ⇒ cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\)

Corollary:

If θ is the angle between the circles x^{2} + y^{2} + 2gx + 2fy + c = 0, x^{2} + y^{2} + 2g’x + 2f’y + c’= 0 then cos θ = \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)

proof:

Let C_{1}, C_{2} be the centre s and r_{1}, r_{2} be the radii of the circles S = 0, S’ = 0 respectively and C_{1}C_{2} = d.

∴ C_{1} = (- g, – f), C_{2} = (- g’, – f’),

r_{1} = \(\sqrt{g^{2}+f^{2}-c}\), r_{2} = \(\sqrt{g^{2}+f^{2}-c}\)

Now cos θ = \(\frac{\mathrm{d}^{2}-\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}}{2 \mathrm{r}_{1} \mathrm{r}_{2}}\) = \(\frac{\left(g-g^{\prime}\right)^{2}+\left(f-f^{\prime}\right)^{2}-\left(g^{2}+f^{2}-c\right)-\left(g^{\prime 2}+f^{\prime 2}-c^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{2}+f^{\prime 2}-c^{\prime}}}\)

= \(\frac{\mathrm{g}^{2}+\mathrm{g}^{\prime 2}-2 \mathrm{gg}^{\prime}+\mathrm{f}^{2}+\mathrm{f}^{\prime 2}-2 \mathrm{ff} \mathrm{f}^{\prime}-\mathrm{g}^{2}-\mathrm{f}^{2}+\mathrm{c}-\mathrm{g}^{2}-\mathrm{f}^{\prime 2}+\mathrm{c}^{\prime}}{2 \sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}} \sqrt{\mathrm{g}^{2}+\mathrm{f}^{\prime 2}-\mathrm{c}^{\prime}}}\)

= \(\frac{c+c^{\prime}-2\left(g g^{\prime}+f f^{\prime}\right)}{2 \sqrt{g^{2}+f^{2}-c} \sqrt{g^{\prime 2}+f^{\prime 2}-c^{\prime}}}\)

Note: Let d be the distance between the centers of two intersecting circles with radii r_{1}, r_{2}. The two circles cut orthogonally if d^{2} = r_{1}^{2} + r_{2}^{2}.

Note: The condition that the two circles

S = x^{2} + y^{2} + 2gx + 2fy + c = 0, S’ = x^{2} + y^{2} + 2g’x + 2f’y + c’ = 0 may cut each other orthogonally is 2gg’ + 2ff’ = c + c’.

Proof: Let C_{1}, C_{2} be the centers and r_{1}, r_{2} be the radii of the circles S = 0, S’ = 0 respectively.

∴ C_{1} = (- g, – f), C_{2} = (- g’, – f’)

r_{1} = \(\sqrt{g^{2}+f^{2}-c}\), r_{2} = \(\sqrt{g^{2}+f^{2}-c^{\prime}}\)

Let P be point of intersection of the circles.

The two circles cut orthogonally at P

⇔ ∠C_{1}PC_{2} = 90°.

⇒ C_{1}C_{2}^{2} = C_{1}P^{2} + C_{2}P^{2} ⇔ (g – g’)^{2} + (f – f’)^{2} = r_{1}^{2} + r_{2}^{2};

⇔ g^{2} + g’^{2} – 2gg’ + f^{2} + f’^{2} – 2ff’ = g^{2} + f^{2} – c + g’^{2} + f’^{2} + c’

⇔ – (2gg’ + 2ff’) = – (c + c’) ⇒ 2gg’+ 2ff’ = c + c’

Note:

- The equation of the common chord of the intersecting circles s = 0 and s
^{1}= 0 is s – s^{1}= 0. - The equation of the common tangent of the touching circles s = 0 and s
^{1}= 0 is s – s^{1}= 0 - If the circle s = 0 and the line L = 0 are intersecting then the equation of the circle passing through the points of intersection of s = 0 and L = 0 is S + λL = 0.
- The equation of the circle passing through the point of intersection of S = 0 and S’ = 0 is s + λS’ = 0.

Theorem: The equation of the radical axis of the circles S = 0, S’ = 0 is S – S’ = 0.

Theorem: The radical axis of two circles is perpendicular to their line of centers.

Proof:

Let S = x^{2} + y^{2} + 2gx + 2fy + c = 0, S’ = x^{2} + y^{2} + 2g’x + 2f’y + c’= 0 be the given circles.

The equation of the radical axis is S – S’ = 0

⇒ 2(g – g’)x + 2(f – f’)y + (c – c’) = 0

⇒ a_{1}x + b_{1}y + c_{1} = 0 where

a_{1} = 2(g – g’), b_{1} = 2(f – f’), c_{1} = e – e’

The centers of the circles are (- g, – f), (- g’, – f’)

The equation to the line of centers is:

(x + g) (f – f’) = (y + f) (g – g’)

⇒ (f – f’)x – (g – g’)y – gf’ + fg’= 0

⇒ a_{2}x + b_{2}y + c_{2} = 0 where

a_{2} = f – f’, b_{2} = – (g – g’), c_{2} = fg’ – gf’

Now a_{1}a_{2} + b_{1}b_{2} = 2(g – g’) (f – f’) – 2(f – f’) (g – g’) = 0.