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}}](https://tex.z-dn.net/?f=x%5E2%20%3D%20%5Cdfrac%7B9%7D%7B%5Cdfrac%7B13%7D%7B4%7D%7D)
![x^2 = {9} \times {\dfrac{4}{13}}](https://tex.z-dn.net/?f=x%5E2%20%3D%20%7B9%7D%20%5Ctimes%20%7B%5Cdfrac%7B4%7D%7B13%7D%7D)
![x= \sqrt{{9} \times {\dfrac{4}{13}}}](https://tex.z-dn.net/?f=x%3D%20%5Csqrt%7B%7B9%7D%20%5Ctimes%20%7B%5Cdfrac%7B4%7D%7B13%7D%7D%7D)
![x= \pm 3 \times \dfrac{2}{\sqrt{13}}}](https://tex.z-dn.net/?f=x%3D%20%5Cpm%203%20%5Ctimes%20%5Cdfrac%7B2%7D%7B%5Csqrt%7B13%7D%7D%7D)
![x= \pm \dfrac{6}{\sqrt{13}}}](https://tex.z-dn.net/?f=x%3D%20%5Cpm%20%5Cdfrac%7B6%7D%7B%5Csqrt%7B13%7D%7D%7D)
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.