Question

Find the points of intersection of the line y = x - 3 and circle (x-3²)² + (y+2)² = 20.

Answer 1

Hint: Here we will use the fact that the points can find out by putting the values from the line in the equation of the circle.

Complete step by step solution:
Given: Equation of line is given as y = x-3 and equation of circle is given as \[{\left( {x - 3} \right)^2} + {\rm{ }}{\left( {y + 2} \right)^2} = {\rm{ }}20\]

Given equation of line is:
 \[y{\rm{ }} = {\rm{ }}x - 3\] …… (i) (Let us consider it as equation (i))
By putting the value of y from equation (i) in the equation of circle, we will get:
\[{\left( {x - 3} \right)^2} + {\rm{ }}{\left( {\left( {x - 3} \right){\rm{ }} + {\rm{ }}2} \right)^2} = {\rm{ }}20\]
\[{\left( {x - 3} \right)^2} + {\rm{ }}{\left( {x{\rm{ }} - {\rm{ }}1} \right)^2} = {\rm{ }}20\]
\[{x^2} + {\rm{ }}9{\rm{ }} - {\rm{ }}6x{\rm{ }} + {\rm{ }}{x^2} + {\rm{ }}1{\rm{ }} - {\rm{ }}2x{\rm{ }} = {\rm{ }}20\]
\[2{x^2} - {\rm{ }}8x{\rm{ }} + {\rm{ }}10{\rm{ }} = {\rm{ }}20\]
\[2{x^2} - {\rm{ }}8x{\rm{ }} - {\rm{ }}10{\rm{ }} = {\rm{ }}0\]
\[{x^2} - {\rm{ }}4x{\rm{ }}-{\rm{ }}5{\rm{ }} = {\rm{ }}0\]
By solving this quadratic equation for x, we will get:
\[\left( {x - 5} \right) \times \left( {x + 1} \right){\rm{ }} = {\rm{ }}0\]
This will give two values of x as:
x = 5 or x = -1
Putting these values of x in equation (i)
y = 5- 3 or y = -1-3
y = 2 or y = -4

Hence intersection points are (5, 2) and (-1, 4)

Note: In such kind of problem we start with finding the x coordinates of the point of intersection according to the problem statement and then put x coordinates in the equation to find y coordinates.