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=2x+5 intersect the circle with radius 5 and center (0, 5)?

Answer 1

Answer:

(2.236,9.472).

Step-by-step explanation:

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$     ...(1)

where, (h,k) is center and r is radius of the circle.

It is given that radius of the circle is 5 and center at (0,5). So, the equation of circle is

$(x-0)^2+(y-5)^2=5^2$

$x^2+(y-5)^2=25$     ...(2)

The given equation of line is

$y=2x+5$       ...(3)

We need to find the intersection point of line and circle in first quadrant.

On solving (2) and (3), we get

$x^2+((2x+5)-5)^2=25$

$x^2+(2x)^2=25$

$x^2+4x^2=25$

$5x^2=25$

$x^2=5$

$x=\pm \sqrt{5}=\pm 2.236$

At x=2.236,

$y=2(2.236)+5=9.472$  

At x=-2.236,

$y=2(-2.236)+5=0.528$  

It means line intersect the circle at (2.236,9.472) and (-2.236,0.528).

In first quadrant both coordinates are positive.

Therefore, the required point is (2.236,9.472).