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 = 2.5 x + 3 intersect the circle with radius 4 and center (0, 3)?

Answer 1

Answer:

The Point in the first quadrant is $(\sqrt[2]{2.0268} ,6.559)$.

Step-by-step explanation:

Since

we know the general equation of the circle whose center is at (h,k) and radius is r that is

$(x-h)^2+(y-k)^2 =r^2$

we have given center is (0,3) and radius is r= 4

substituting in the general equation of the circle we get

$x^2 +(y-3)^2 = 16$

Now substituting the y= 2.5x + 3 in the above equation we get

$x^2 +(2.5x + 3-3)^2 = 4^2$

$x^2 +6.25x^2 = 16$

here subtracting 16 from both sides  and also perform addition

we get

$7.25x^2 = 16$

dividing both sides by 7.25 we get

$x^2 =2.2068$

taking square root both sides we get

$x= \frac{+}{-}\sqrt[2]{2.0268}$

but we need positive value of x because both x and y are positive in the first quadrant

that is

$x= \sqrt[2]{2.0268}$

substituting this into the given equation of the line we get

y =  2.5($\sqrt[2]{2.0268}$)+3

y =  6.559

so the point in the first quadrant is ($\sqrt[2]{2.0268}$,6.559)

that is required    

Answer 2

Answer:

The point at which the line y  = 2.5·x + 3 intersect with the circle of radius 4 and center (0, 3) is x = 1.49

Step-by-step explanation:

Here we have the equation of a circle = (x - h)² + (y-k)² = r²

Where (h, k) is the coordinate of the center

Therefore, we have

(x - 0)² + (y - 3)² = 4²

Which gives x² + (y - 3)² = 16

Since the line y  = 2.5·x + 3 we have

x² + ( (2.5·x + 3 ) - 3)² = 16

x² + (2.5·x)² = 16

x² + 6.25·x² = 16

7.25·x² = 16

x² = 16/7.25 = 2.21

x = 1.49

That is the line y  = 2.5·x + 3 intersect with the circle of radius 4 and center (0, 3) at x = 1.49.