Question

Recall the equation for a circle with center (h,k) and radius r. At what point in the first quadrant does the line with equation y
=2.5x+2 intersect the circle with radius 4 and center (0, 2)?

Answer 1

Answer:

At point (1.486, 5.714)

Step-by-step explanation:

The equation of the circle with radius 4 and center at (0, 2) is:

$x^2+(y-2)^2=16$

and we want to know where in the 1st quadrant does it intersect with line $y=2.5x+2$

To find the point of intersection we put the value of  $y=2.5x+2$ into the equation of the circle:

$x^2+((2.5x+2)-2)^2=16$

$x^2+(2.5x)^2=16$

$7.25x^2=16$

$x=\pm 1.486$

and since we only concerned with the 1st quadrant we take the positive value of $x=1.486$ which is the x coordinate of intersection.

The y coordinate is found by putting this x value into either the circle equation or the line equation ; we choose the line equation $y=2.5x+2$ because it is easier to work with.

$y=2.5(1.486)+2=5.714$

$\boxed{y=5.714}$

Thus we have the point of intersection

$\boxed{(x,y)=(1.486, 5.714)}$