Question

How do you find the intersection between \[y = x - 8\] and \[{x^2} + {y^2} = 34\]?

Answer 1

Hint: We are given with a pair of equations. One of the equations is of a straight line and the other is of a circle with origin as the center. We will use a method of substitution to find the points of intersection. We will convert the equation of the circle in the form of y and will equate with the equation of the line that is also in the form of y. Then on solving we will get the values of y. With the help of that we will find the value of x also. And so the points of intersection.

Complete step-by-step answer:
Given is,
\[y = x - 8\]…..equation of a line (1)
\[{x^2} + {y^2} = 34\]….equation of circle (2)
We can modify the equation of circle as in the form of y,
\[{y^2} = 34 - {x^2}\]
Taking the roots on both the sides,
\[y = \pm \sqrt {34 - {x^2}} \]…..equation(3)
Equating equation 1 and 3 because their LHS are same,
\[x - 8 = \pm \sqrt {34 - {x^2}} \]
Squaring both sides in order to remove the root,
\[{\left( {x - 8} \right)^2} = 34 - {x^2}\]
Taking the square,
\[{x^2} - 2 \times x \times 8 + {8^2} = 34 - {x^2}\]
On solving further,
\[{x^2} - 16x + 64 = 34 - {x^2}\]
Rearranging the terms,
\[{x^2} + {x^2} - 16x + 64 - 34 = 0\]
\[2{x^2} - 16x + 30 = 0\]
On dividing the whole equation by 2,
\[{x^2} - 8x + 15 = 0\]
This is a quadratic equation of the form \[a{x^2} + bx + c = 0\]. We will solve the equation to find the factors. The factors are the numbers Such that the middle term should be the sum and last term should be the product.
\[{x^2} - 5x - 3x + 15 = 0\]
Taking the terms common from first two and last two terms,
\[x\left( {x - 5} \right) - 3\left( {x - 5} \right) = 0\]
Separating the brackets,
\[\left( {x - 3} \right)\left( {x - 5} \right) = 0\]
Now we will tabulate the values of both x and y.

Value of x	Value of y
\[ x - 3 = 0 \\ x = 3 \\ \]	Using equation of line, \[\begin{gathered} y = x - 8 \\ y = 3 - 8 \\ y = - 5 \\ \end{gathered} \]
\[ x - 5 = 0 \\ x = 5 \\ \]	Using equation of line,\[\begin{gathered} y = x - 8 \\ y = 5 - 8 \\ y = - 3 \\ \end{gathered} \]

Thus the points of intersection are \[\left( {3, - 5} \right)\& \left( {5, - 3} \right)\].
So, the correct answer is “ \[\left( {3, - 5} \right)\& \left( {5, - 3} \right)\].”.

Note: Here note that the points of intersection are nothing but the points where the line and the circle intersect each other. This means that both equations should be satisfied by the points of intersection. So we can use the method of equating the equations as we did in the problem above.