Question

Find the point of Intersection of the circle $x^2+y^2=4$ with line passing through $A(1,0)$ and $B(3,4)$.

Answer 1

Circle : $x^2+y^2=4\Rightarrow$ Centre $=(0,0)$ Radius $2$ units

line passing through: $A(1,0)$ and $B(3,4)$

Equation of line passing through two given points

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$(x_1,y_1)=(1,0)$

$(x_2,y_2)=(3,4)$

$\Rightarrow y-0=\frac{4-0}{3-1}(x-x_1)$

$\Rightarrow y=2(x-1)$ Equation of given line

At the point of intersection both line and circle will passing through the same point $\therefore$ Putting $y=2(x-1)$ in the equation of the circle.

$x^2+\left[2\right(x-1){]}^2=4\Rightarrow x^2+4(x^2+1-2x)=4$

$\Rightarrow x^2+4x^2+4-8x=4\Rightarrow 5x^2-8x=0$

$\Rightarrow x(5x-8)=0\Rightarrow x=0$ or $x=8/5$

Putting $x=0$ and $x=8/5$ in the equation of the line to get respective values of $y$

For $x=0,y=2(0-1)=-2$ $\frac{}{}]\Rightarrow$ Point of intersection are $(0,-2)$ and $(8/5,6/5)$

For $x=8/5,y=2(8/5-1)=6/5$

Points of intersection : $(0,-2)$ and $(8/5,6/5)$