Answer :
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
[tex]x^2 = \dfrac{9}{\dfrac{13}{4}}[/tex]
[tex]x^2 = {9} \times {\dfrac{4}{13}}[/tex]
[tex]x= \sqrt{{9} \times {\dfrac{4}{13}}}[/tex]
[tex]x= \pm 3 \times \dfrac{2}{\sqrt{13}}}[/tex]
[tex]x= \pm \dfrac{6}{\sqrt{13}}}[/tex]
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.