Question

Recall the question for a circle with center. (H,k) and Radius r. At what point in the first quadrant does the line with equation y=1.5x+1 intersect with radius 3 and center (0,1)

Answer 1

Answer:

the point lies at (1.7, 3.6) in the first quadrant.

Step-by-step explanation:

The formula for a circle with centre (h,k) and radius (r) can be expressed as :

(x-h)² + (y-k)² = r²

Now for the expression of the circle with radius 3 and center (0,1), we have:

(x - 0)² + (y - 1)² = 3²

x² + (y-1)² = 9

replacing y = 1.5x + 1 in the above equation, we have:

x² + (( 1.5x + 1 ) - 1 )² = 9

x² + ( 1.5x +1 - 1)² = 9

x² + (1.5x)² = 9

x² + 2.25x² = 9

3.25x² = 9

x² = 9/3.25

$x^2 = \dfrac{9}{\dfrac{13}{4}}$

$x^2 = {9} \times {\dfrac{4}{13}}$

$x= \sqrt{{9} \times {\dfrac{4}{13}}}$

$x= \pm 3 \times \dfrac{2}{\sqrt{13}}}$

$x= \pm \dfrac{6}{\sqrt{13}}}$

x = 1.7 in the positive x - axis

Recall that:

y = 1.5x + 1

y = 1.5 (1.7) +1

y = 2.55 +1

y = 3.55

y = 3.6

Therefore, the point lies at (1.7, 3.6) in the first quadrant.