Question

= 10 The line 7x + 2y = –20 intersects the curve x2 + y2 + 4x + 6y – 40 = 0 at the points A and B. Find the length of the line AB.​

Answer 1

$\Huge2\sqrt{53}$

$\huge\textbf{Steps to solution}$

$\Large\textrm{Given question is: -}$

$\textrm{The line 7x+2y=-20 intersects the curve}$

$\textrm{ x^{2}+y^{2}+4x+6y-40=0 at the points A and B.}$

$\textrm{ \rightarrow Find the length of \overline{\textrm{AB}} .}$

$\Large\textrm{Here is the solution: -}$

$\textrm{We know that d=\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}} is given as}$

$\textrm{the distance from a point to a line.}$

$\textrm{We have: -}$

$\textrm{ \rightarrow a=7 , b=2 , c=20 , x_{1}=-2 , y_{1}=-3 }$

$\textrm{The distance is given as: -}$

$d=\dfrac{|-14-6+20|}{\sqrt{49+4}}$

$d=\dfrac{0}{\sqrt{53}}$

$d=0$

$\textrm{Hence, the line segment \overline{\textrm{AB}} is the diameter.}$

$\textrm{ x^{2}+y^{2}+4x+6y-40=0 }$

$(x^{2}+4x+4)-4+(y^{2}+6y+9)-9-40=0$

$(x+2)^{2}+(y+3)^{2}-53=0$

$(x+2)^{2}+(y+3)^{2}=53$

$\textrm{Hence, the diameter is 2r=2\sqrt{53} .}$

$\huge\textbf{Formulae used: -}$

$\Large\textrm{ \bigstar Distance to Line \bigstar }$

$\boxed{d=\dfrac{|ax_{1}+by_{1}+c|}{\sqrt{a^{2}+b^{2}}}}$

$\Large\textrm{ \bigstar Distance to Circle \bigstar }$

$\textrm{ d<r then the line meets the circle twice.}$

$\textrm{ d=r then the line is tangent to the circle.}$

$\textrm{ d>r then the line does not meet the circle.}$