Question

The equation of the tangent to a circle at P(2,4) is 3x = x+2. The line y = 3x passes through the center C of the circle. Finda) the coordinates of C.b)the equation of the circle.c) the point T1 and T2 where the circle cuts the y-axis.

Answer 1

ANSWER and EXPLANATION

(a) To find the coordinates of point C, the center of the circle, we have to first find the equation of the line CP.

Since the given line:

`3y=x+2`

is a tangent to CP, it implies that its slope is the negative inverse of the slope of line CP.

Let us put the equation in standard form:

`y=(1)/(3)x+(2)/(3)`

Hence, its slope is 1/3.

Therefore, the slope of CP is:

`m=-3`

Since the line CP touches the point P, we can find the equation of the line using the point-slope method:

`y-y_1=m(x-x_1)`

where (x1, y1) = point the line passes through

m = slope

Hence, the equation of line CP is:

`\begin{gathered} y-2=-3(x-4) \\ y-2=-3x+12 \\ y=-3x+12+2 \\ y=-3x+14 \end{gathered}`

The equation above and the equation y = 3x intersect at point C. This implies that point C is the solution to the two equations.

We can find the x-coordinate of C by equating the equations:

`\begin{gathered} 3x=-3x+14 \\ 3x+3x=14 \\ 6x=14 \\ x=(14)/(6)=(7)/(3) \end{gathered}`

Now, find the y-coordinate by substituting the x value into either equation:

`\begin{gathered} y=3((7)/(3)) \\ y=7 \end{gathered}`

Hence, the coordinates of C are:

`((7)/(3),7)`

(b) The general form for the equation of a circle is given by:

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

where (h, k) = center of the circle

r = radius of the circle

C is the center of the circle, which implies that:

`(h,k)=((7)/(3),7)`

To find the radius of the circle, find the length of CP using the formula:

`CP=√((x_2-x_1)^2+(y_2-y_1)^2)`

where (x1, y1) and (x2, y2) are the coordinates of C and P.

Hence, the radius of the circle is:

`\begin{gathered} r=\sqrt{(4-(7)/(3))^2+(2-7)^2}=\sqrt{((12)/(7))^2+(-5)^2} \\ r=\sqrt{(144)/(49)+25}=\sqrt{(1369)/(49)} \\ r=(37)/(7) \end{gathered}`

Therefore, the equation of the circle is:

`(x-(7)/(3))^2+(y-7)^2=((37)/(7))^2`

(c) To find the points where the circle cuts the y-axis, solve for y in the equation above when x is 0:

`\begin{gathered} (0-(7)/(3))^2+(y-7)^2=((37)/(7))^2 \\ \\ (y-7)^2=(1369)/(49)-(49)/(9) \\ \\ (y-7)^2=(9920)/(441) \\ \\ y^2-14y+49=(9920)/(441) \\ \\ y^2-14y+49-(9920)/(441)=0 \\ \\ y^2-14y+(11689)/(441)=0 \end{gathered}`

Now, solve the equation above using the quadratic equation:

`y=(-b\pm√(b^2-4ac))/(2a)`

where a = 1, b = -14, c = 11689/441

Therefore, the solution is:

`\begin{gathered} y=\frac{14\pm\sqrt{(-14)^2-(4*1*(11689)/(441))}}{2(1)} \\ y=\frac{14\pm\sqrt{196-(46756)/(441)}}{2}=\frac{14\pm\sqrt{(39680)/(441)}}{2} \\ y=(14\pm9.49)/(2) \\ y=(14+9.49)/(2);y=(14-9.49)/(2) \\ y=11.74;y=2.26 \end{gathered}`

Hence, the coordinates of T1 and T2 are:

`\begin{gathered} T_1=(0,2.26) \\ T_2=(0,11.74) \end{gathered}`