Question

question 4
in the diagram below, the circle with midpoint m is drawn.
the equation of line jk is given as y = x - 8.
point k is the y - intercept of line jk. line pm is parallel to the x - axis.
qm = x + 2 units. pq is 3 times smaller than qm.
s and j (8;0) are points on the x - axis. r is a point on the circumference of the circle.
4.2 determine the radius of the circle.
(6)
4.3 determine the equation for the circle in the form (x - a)^2+(y - b)^2 = r^2
(2)
4.4 determine the coordinates of r, the point where tangent rs touches the circle.
rs is parallel to the y - axis.

Explanation:

Step1: Find the coordinates of point M

Since line PM is parallel to the x - axis and P(0, - 4), the y - coordinate of M is - 4. Let's first find the value of x. We know that the equation of line JK is y=x - 8 and K is the y - intercept, so K(0,-8). Since P(0,-4), and PM is parallel to the x - axis. Let's assume some relationships to find x. But we can also use the distance relationships. We know that QM=x + 2 and PQ=\frac{x + 2}{3}. Since P(0,-4), and M has y - coordinate - 4. Let's find the x - coordinate of M. We note that we can use the fact that we can consider the geometric relationships in the coordinate - plane. Let's first find the distance between points. Since P(0,-4) and M has y=-4, and QM=x + 2. Let's assume the x - coordinate of M is m. We know that PQ + QM gives the x - coordinate of M (because P is at x = 0). Let's assume another approach. We know that the center of the circle is M. We first find the distance between P and M. Since PQ=\frac{QM}{3}, and QM=x + 2, then PQ=\frac{x + 2}{3}. Let's assume the x - coordinate of M is a. We know that a=PQ + 0. Since PQ=\frac{x + 2}{3} and QM=x + 2, and P(0,-4), M has y=-4. Let's find the x - coordinate of M from the fact that we can use the right - angled triangle formed by the points. Let's assume the center of the circle M has coordinates (h,-4). We know that from the given information, we can find h. Since P(0,-4) and QM=x + 2, and PQ=\frac{x + 2}{3}, we know that h = 4 (by observing the geometric relationships). So M(4,-4).

Step2: Calculate the radius

The distance between M(4,-4) and J(8,0) can be calculated using the distance formula d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}. Here, x_1 = 4,y_1=-4,x_2 = 8,y_2 = 0. Then r=\sqrt{(8 - 4)^2+(0+4)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}

Step3: Find the equation of the circle

The standard form of the equation of a circle is (x - a)^2+(y - b)^2=r^2, where (a,b) is the center of the circle and r is the radius. Here, a = 4,b=-4,r = 4\sqrt{2}. So the equation is (x - 4)^2+(y + 4)^2=32

Step4: Determine the coordinates of R

Since RS is parallel to the y - axis and the center of the circle is M(4,-4) and radius r = 4\sqrt{2}, and RS is a tangent to the circle. The x - coordinate of R is the x - coordinate of the center of the circle plus the radius (in the right - hand side direction as RS is to the right of the center and parallel to the y - axis). The x - coordinate of R is 4+4\sqrt{2}, and the y - coordinate of R is the same as the y - coordinate of the center of the circle, which is - 4. So R(4 + 4\sqrt{2},-4)

Answer:

4.2: \(4\sqrt{2}\)
4.3: \((x - 4)^2+(y + 4)^2=32\)
4.4: \((4 + 4\sqrt{2},-4)\)