Circles
7.0 Length of intercept cutoff from a line by a circle or length of chord of a circle
7.0 Length of intercept cutoff from a line by a circle or length of chord of a circle
Consider standard form of circle ${x^2} + {y^2} = {a^2}$ and a line $y=mx+c$ intersects the circle at two points $P({x_1},{y_1})$ and $Q({x_2},{y_2})$.
Points $P$ and $Q$ lie on the line $y=mx+c$ and satisfies the equation, so, $${y_1} = m{x_1} + c...(1)$$ and $${y_2} = m{x_2} + c...(2)$$ Subtract equation $(1)$ from $(2)$, we get $${y_2} - {y_1} = m\left( {{x_2} - {x_1}} \right)...(3)$$ Distance between points $P$ and $Q$ using distance formulae, $$PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} ...(4)$$
Put the value of ${y_2} - {y_1}$ from equation $(3)$ to equation $(4)$ $$PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {m^2}{{\left( {{x_2} - {x_1}} \right)}^2}} $$ $$ = \left| {{x_2} - {x_1}} \right|\sqrt {1 + {m^2}} $$ $$ = \sqrt {{{\left( {{x_2} + {x_1}} \right)}^2} - 4{x_2}{x_1}} \times \sqrt {1 + {m^2}} ...(5)$$
Substitute $y=mx+c$ in equation of circle $${x^2} + {\left( {mx + c} \right)^2} = {a^2}$$ $${x^2}\left( {1 + {m^2}} \right) + 2mcx + {c^2} - {a^2} = 0$$ The above equation has roots ${x_1}$ and ${x_2}$ Therefore, $${x_1} + {x_2} = \frac{{ - 2mc}}{{1 + {m^2}}}$$ and $${x_1} \times {x_2} = \frac{{{c^2} - {a^2}}}{{1 + {m^2}}}$$
Put these values in equation $(5)$, we get $PQ = \sqrt {\frac{{4{m^2}{c^2}}}{{{{\left( {1 + {m^2}} \right)}^2}}} - \frac{{4\left( {{c^2} - {a^2}} \right)}}{{1 + {m^2}}}} \times \sqrt {1 + {m^2}} $
$$PQ\left( {length\ of\ intercept} \right) = 2\sqrt {\frac{{{a^2}\left( {1 + {m^2}} \right) - {c^2}}}{{1 + {m^2}}}} $$
