Question

section b 10 marks
answer one question from this section
1 find the length of the tangent from the point (5,7) to the circle
( x^2 + y^2 - 4x - 6y + 9 = 0 )
2 find the equation of the parabola having its vertex at the origin and symmetrical about the x-axis and passing through the point (-4, -10). sketch its graph.
3 find the coordinates of the center and the radius of the circle with equation
( x^2 + y^2 + 6x + 12y + 9 = 0 )

Explanation:

Question 1: Find the length of the tangent from the point \((5,7)\) to the circle \(x^{2}+y^{2}-4x - 6y + 9 = 0\)

Step 1: Recall the formula for the length of the tangent

The length of the tangent from a point \((x_1,y_1)\) to the circle \(x^{2}+y^{2}+2gx + 2fy + c = 0\) is given by \(\sqrt{x_1^{2}+y_1^{2}+2gx_1 + 2fy_1 + c}\)

First, rewrite the given circle equation \(x^{2}+y^{2}-4x - 6y + 9 = 0\) in the form \(x^{2}+y^{2}+2gx + 2fy + c = 0\). Here, \(2g=-4\) so \(g = - 2\), \(2f=-6\) so \(f=-3\) and \(c = 9\)

The point is \((x_1,y_1)=(5,7)\)

Step 2: Substitute the values into the formula

Substitute \(x_1 = 5\), \(y_1=7\), \(g=-2\), \(f = - 3\) and \(c = 9\) into the formula for the length of the tangent \(L=\sqrt{x_1^{2}+y_1^{2}+2gx_1+2fy_1 + c}\)

\[

$$\begin{align*} L&=\sqrt{5^{2}+7^{2}+2\times(-2)\times5+2\times(-3)\times7 + 9}\\ &=\sqrt{25 + 49-20-42 + 9}\\ &=\sqrt{(25 + 49+9)-(20 + 42)}\\ &=\sqrt{83 - 62}\\ &=\sqrt{21} \end{align*}$$

\]

Step 1: Recall the standard equation of a parabola

The standard equation of a parabola with vertex at the origin and symmetric about the \(x\)-axis is \(y^{2}=4ax\) (if it opens to the right) or \(y^{2}=- 4ax\) (if it opens to the left). Since the point \((-4,-10)\) has \(x=-4<0\) and \(y = - 10\), we consider the form \(y^{2}=-4ax\) (opens to the left)

Step 2: Substitute the point into the equation

The point \((x,y)=(-4,-10)\) lies on the parabola. Substitute \(x=-4\) and \(y=-10\) into \(y^{2}=-4ax\)

\[

$$\begin{align*} (-10)^{2}&=-4a\times(-4)\\ 100&=16a\\ a&=\frac{100}{16}=\frac{25}{4} \end{align*}$$

\]

Step 3: Write the equation of the parabola

Substitute \(a = \frac{25}{4}\) into \(y^{2}=-4ax\)

\[

$$\begin{align*} y^{2}&=-4\times\frac{25}{4}x\\ y^{2}&=-25x \end{align*}$$

\]

We can verify by substituting \((-4,-10)\) into \(y^{2}=-25x\): \((-10)^{2}=100\) and \(-25\times(-4) = 100\), so it satisfies.

Step 1: Recall the standard form of a circle

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. To convert the given equation \(x^{2}+y^{2}+6x + 12y + 9 = 0\) to standard form, we complete the square.

Step 2: Complete the square for \(x\) and \(y\) terms

For the \(x\)-terms: \(x^{2}+6x=(x + 3)^{2}-9\)

For the \(y\)-terms: \(y^{2}+12y=(y + 6)^{2}-36\)

Substitute these into the given equation:

\[

$$\begin{align*} (x + 3)^{2}-9+(y + 6)^{2}-36+9&=0\\ (x + 3)^{2}+(y + 6)^{2}-36&=0\\ (x + 3)^{2}+(y + 6)^{2}&=36 \end{align*}$$

\]

Step 3: Identify the center and radius

Comparing with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-3\), \(k = - 6\) and \(r^{2}=36\) so \(r = 6\)

Answer:

\(\sqrt{21}\)

Question 2: Find the equation of the parabola having its vertex at the origin and symmetrical about the \(x\)-axis and passing through the point \((-4,-10)\)