## Intersection of a Line and a Circle:

Let us consider the line y = mx + c and the circle x^{2} + y^{2} = r^{2} (whose center is at the origin). For the point of intersection of the line and the circle, we have

x^{2} + (mx + c)^{2} = r^{2}⇒ x ^{2} + m^{2}x^{2} + 2mcx + c^{2} – r^{2} = 0⇒ (1 + m ^{2}) x^{2} + 2mcx + (c^{2} – r^{2}) = 0 …………(i)which is a quadratic in x having two roots x _{1} and x_{2}. The point of intersection are (x_{1}, mx_{1} + c) and (x_{2}, mx_{2} + c).The discriminant of the equation is (2mc) ^{2} – 4(1 + m^{2}) (c^{2} – r^{2}) = 4m^{2}c^{2} – 4(c^{2} + m^{2}c^{2} – r^{2} – r^{2}m^{2}) = 4 { r^{2} (1+ m^{2}) – c^{2} } = D |

(i) If c^{2} < r^{2}(1 + m^{2}), then D > 0. The roots x_{1} and x_{2} are then real and distinct. In this case, the line y = mx + c intersects the circle x^{2} + y^{2} = r^{2} at two distinct points A and B as shown in the figure below. Here, r > | c / √(1 + m^{2}) | i.e., the radius of the circle is more than the distance of the line from the center.

(ii) If c^{2} = r^{2}(1 + m^{2}), then D = 0. The roots x_{1} and x_{2} of equation (i) are equal. The line y = mx + c, in this case, intersects the circle x^{2} + y^{2} = a^{2} at two real and coincident points as shown in the figure below. Here, r = | c / √(1 + m^{2}) i.e., the radius of the circle is equal to the distance of the line from the center.

(iii) If c^{2} > r^{2}(1 + m^{2}), then D < 0. The roots x_{1} and x_{2} of equation (i) are imaginary. The line y = mx + c does not intersect the circle as shown in the figure below. In this case, r < | c / √(1 + m^{2}) | i.e., the radius of the circle is less than the distance of the line from the center of the circle.

## Length of the Intercepted Chord:

Let the line y = mx + c intersect the circle at two points A and B as shown in the figure. If the coordinates of points A and B be (x_{1}, y_{1}) and (x_{2}, y_{2}) respectively. The length of the chord is AB.

For A and B, x ^{2} + (mx + c)^{2} – r^{2} = 0 ⇒ (1 + m ^{2}) x^{2} + 2mcx + (c^{2} – r^{2}) = 0As x _{1}, x_{2} are the roots of equation (i), we havex _{1} + x_{2} = -2mc / (1 + m^{2}) and x_{1}x_{2} = (c^{2} – r^{2}) / (1 + m^{2}) Now, A and B are points on y = mx + c ∴ y _{1} = mx_{1} + c and y_{2} = mx_{2}+ c∴ y _{1} – y_{2} = m (x_{1} – x_{2})Now, the length of the Chord AB = √[ (x_{1} – x_{2})^{2} + (y_{1} – y_{2})^{2} ] = √[ (x_{1} – x_{2})^{2} + m^{2 }(x_{1} – x_{2})^{2} ]= √[ (x _{1} – x_{2})^{2} (1 + m^{2}) ]= √[ { (x _{1} + x_{2})^{2} – 4x_{1}x_{2} } (1 + m^{2}) ] |

*Note:*

If AB = 0, then r^{2}(1 + m^{2}) – c^{2} = 0⇒ c = ± r √(1 + m ^{2}). In this case, A and B coincide and the line y = mx + c becomes a tangent to the circle x^{2} + y^{2} = r^{2}. Thus, the condition of tangency of the line y = mx + c to the circle x^{2} + y^{2} = r^{2} is c = ± r √(1 + m^{2}) or, r = | c / √(1 + m^{2}) | i.e., the radius of the circle becomes equal to the length of the p[erendicular drawn from the center (0, 0) to the line y = mx + c. |

Find the coordinates of the point of intersection of the line x + √3 y + 7 = 0 and the circle xExample 1:^{2} + y^{2} =10. The given equation of a line is,Solution-x + √3 y + 7 = 0 ⇒ x = -7 – √3 y ……….(i) and the equation of a circle is, x ^{2} + y^{2} =10⇒ (-7 – √3 y) ^{2} + y^{2} = 10⇒ 49 + 3y ^{2} + 14√3 y + y^{2} = 10⇒ 4y ^{2} + 14√3 y + 39 = 0⇒ y = [ -14√3 ± √{(14√3) ^{2} – 4(4) (39)} ] / 2(4) ⇒ y = [ -14√3 ± √(588 – 624) ] / 8 ⇒ y = [ -14√3 ± 6i ] / 8 which gives imaginary values of y ∴ The line and circle do not intersect each other. |

For what value of c is the line y = 2x + c tangent to the circle xExample 2:^{2} + y^{2} = 5? The given equation of a line is,Solution-y = 2x + c ⇒ 2x – y + c = 0 ………..(i) and the equation of a circle is, x ^{2} + y^{2} = 5⇒ x ^{2} + y^{2} = (√5)^{2}Center = (0, 0) Radius = √5 Since the line is tangent to the circle. The length of ⊥ from (0, 0) to the line, 2x – y + c = 0 is equal to the radius of a circle. ⇒ | 2(0) – 0 + c | / √[(2) ^{2} + (1)^{2}] = √5⇒ | c | / √5 = √5 ⇒ | c | = 5 ⇒ c = ± 5 |

Find the value of k for which the line x + y + k = 0 touches the circle xExample 3:^{2} + y^{2} – 2x – 4y + 3 = 0. The given equation is,Solution-x + y + k = 0 and the equation of a circle is, x ^{2} + y^{2} – 2x – 4y + 3 = 0Center = (1, 2) and Radius (r) = √[ (1) ^{2} + (2)^{2} – 3 ] = √2Since the line touches the circle. ∴ The length of ⊥ from the point (1, 2) to the line x + y + k = 0 is equal to the radius of a circle. ⇒ | 1 + 2 + k | / √[ (1) ^{2} + (1)^{2} ] = √2⇒ | 3 + k | = 2 ⇒ 3 + k = ±2 ⇒ 3 + k = 2 or 3 + k = -2 ⇒ k = -1 or k = -5 |

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