233k views
2 votes
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 = 1.5x + 3 intersect the circle with radius 6 and center (0,3)

User Wu Zhou
by
8.3k points

2 Answers

6 votes

Answer:


\textsf{Exact point of intersection:}\quad \left((12√(13))/(13),(39+18√(13))/(13)\right)


\textsf{Rounded point of intersection:} \quad (3.328, 7.992)

Explanation:

The equation of a circle is given by:


\large\boxed{(x - h)^2 + (y - k)^2 = r^2}

where:

  • (h, k) is the center.
  • r is the radius.

To find the point(s) of intersection of the line y = 1.5x + 3 and a circle with a center with radius 6 and center (0, 3), substitute the following values into the equation of a circle formula:

  • h = 0
  • k = 3
  • r = 6
  • y = 1.5x + 3

Therefore:


(x - 0)^2 + (1.5x + 3 - 3)^2 = 6^2

Solve for x:


\begin{aligned}(x - 0)^2 + (1.5x + 3 - 3)^2&= 6^2\\\\x^2+(1.5x)^2&=36\\\\x^2+2.25x^2&=36\\\\3.25x^2&=36\\\\(13)/(4)x^2&=36\\\\x^2&=(144)/(13)\\\\x&=\pm \sqrt{(144)/(13)\\\\x&=\pm(12)/(√(13)\\\\x&=\pm(12√(13))/(13))\end{aligned}

As the point is in the first quadrant, we take the positive value of x only:


x=(12√(13))/(13)

To find the y-coordinate of the point, substitute the found value of x into the equation of the line and solve for y:


\begin{aligned}y&=1.5\left((12√(13))/(13)\right)+3\\\\y&=(3)/(2)\left((12√(13))/(13)\right)+3\\\\y&=(36√(13))/(26)+3\\\\y&=(18√(13))/(13)+(39)/(13)\\\\y&=(39+18√(13))/(13)\end{aligned}

Therefore, the point of intersection of the line y = 1.5x + 3 and the circle with center (0, 3) and radius 6 in the first quadrant is exactly:


\left((12√(13))/(13),(39+18√(13))/(13)\right)

If we round the coordinates to the nearest thousandth (3 decimal places), we get:


(3.328, 7.992)

User Jennybryan
by
8.0k points
3 votes

Answer:

The point in the first quadrant is (3.33, 7.995).

Explanation:

The equation for a circle with center (h, k) and radius r is:


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

In this case, the center of the circle is (0, 3) and the radius is 6.

So the equation of the circle is:


\sf (x - 0)^2 + (y - 3)^2 = 6^2

We can rewrite this equation as:


\sf x^2 + y^2 - 6y +9 = 36


\sf x^2 + y^2 - 6y = 36-9


\sf x^2 + y^2 - 6y = 27

The line with equation y = 1.5x + 3 intersects the circle when the two equations are equal.

So we can substitute y = 1.5x + 3 into the equation for the circle to get:


\sf x^2 + (1.5x + 3)^2 - 6(1.5x + 3) = 27

Expand the square term:


\sf x^2 + 2.25x^2 + 9x + 9 - 6(1.5x + 3) = 27

Distribute the constant term 6 on the right side of the equation:


\sf x^2 + 2.25x^2 + 9x + 9 - 9x -18 = 27

Combine like terms:


\sf \sf 3.25x^2 -9+9 = 27+9


\sf 3.25x^2 = 36

Divide both sides by 3.25:


\sf x^2 =(36)/(3.25)

Take the square root of both sides:


\sf x =\sqrt{(36)/(3.25)}


\sf x \approx \pm 3.33

Since we're looking for the point in the first quadrant, take the positive square root:


\sf x \approx 3.33

Now substitute the x value back into the line equation to find the corresponding y value:


\sf y = 1.5* (3.33) + 3


\sf y \approx 4.995 + 3


\sf y \approx 7.995

So, the point of intersection between the line(y = 1.5x + 3 and the circle with center (0, 3)and radius 6 in the first quadrant is approximately (3.33, 7.995).

User Tamil
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories