In the diagram below, a circle with the centre at A (−3; −1) and a tangent CD to the circle at point B. Given…

Question

In the diagram below, a circle with the centre at A (−3; −1) and a tangent CD to the circle at point B. Given also that CD cuts the x-axis at −6 and that the equation of the line segment AB (radius) is −x + 2y − 1 = 0.

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Question 1.1

1.1.1

Determine the relationship between line segments CD and AB.

Identify the given information.
AB is a radius of the circle, and CD is a tangent to the circle at point B.
Recall the geometric property that a radius drawn to the point of tangency is perpendicular to the tangent.
The relationship is that AB is perpendicular to CD.
$\boxed{\text{AB} \perp \text{CD}}$
1.1.2

Determine the equation of the tangent CD.
Find the gradient of the line segment AB.
The equation of AB is $-x + 2y - 1 = 0$ .
Rearrange to $y = mx + c$ form:
$2y = x + 1$ $y = \frac{1}{2}x + \frac{1}{2}$
The gradient of AB is $m_{AB} = \frac{1}{2}$ .
Determine the gradient of the tangent CD.
Since CD is perpendicular to AB, the product of their gradients is $-1$ .
$m_{CD} \cdot m_{AB} = -1$ $m_{CD} \cdot \frac{1}{2} = -1$ $m_{CD} = -2$
Use the point C $(-6, 0)$ (where CD cuts the $x$ -axis at $-6$ ) and the gradient $m_{CD} = -2$ to find the equation of CD using the point-slope form $y - y_1 = m(x - x_1)$ .
$y - 0 = -2(x - (-6))$ $y = -2(x + 6)$ $y = -2x - 12$
The equation of the tangent CD is $\boxed{y = -2x - 12}$ .

1.1.3

Determine the coordinates of point B.
Point B is the intersection of line AB and line CD.
We have their equations:
Equation of AB: $-x + 2y - 1 = 0$ (1)
Equation of CD: $y = -2x - 12$ (2)
Substitute equation (2) into equation (1) to solve for $x$ .
$-x + 2(-2x - 12) - 1 = 0$ $-x - 4x - 24 - 1 = 0$ $-5x - 25 = 0$ $-5x = 25$ $x = -5$
Substitute $x = -5$ into equation (2) to solve for $y$ .
$y = -2(-5) - 12$ $y = 10 - 12$ $y = -2$
The coordinates of point B are $\boxed{(-5, -2)}$ .

1.1.4

Determine the equation of the circle in the form $(x - a)^2 + (y - b)^2 = r^2$ .
Identify the center of the circle.
The center is A $(-3, -1)$ . So, $a = -3$ and $b = -1$ .
Calculate the radius $r$ .
The radius is the distance between the center A $(-3, -1)$ and point B $(-5, -2)$ (a point on the circle). Use the distance formula $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .
$r = \sqrt{(-5 - (-3))^2 + (-2 - (-1))^2}$ $r = \sqrt{(-5 + 3)^2 + (-2 + 1)^2}$ $r = \sqrt{(-2)^2 + (-1)^2}$ $r = \sqrt{4 + 1}$ $r = \sqrt{5}$
Square the radius to get $r^2$ .
$r^2 = (\sqrt{5})^2 = 5$
Substitute the center coordinates and $r^2$ into the standard equation of a circle.
$(x - (-3))^2 + (y - (-1))^2 = 5$ $(x + 3)^2 + (y + 1)^2 = 5$
The equation of the circle is $\boxed{(x + 3)^2 + (y + 1)^2 = 5}$ .

In the diagram below, a circle with the centre at A (−3; −1) and a tangent CD to the circle at point B. Given also that CD cuts the x-axis at −6 and that the equation of the line segment AB (radius) is −x + 2y − 1 = 0.

Step-by-step solution

Question 1.1

1.1.1

Identify the given information.

Recall the geometric property that a radius drawn to the point of tangency is perpendicular to the tangent.

1.1.2

Find the gradient of the line segment AB.

Determine the gradient of the tangent CD.

Use the point C $(-6, 0)$ (where CD cuts the $x$ -axis at $-6$ ) and the gradient $m_{CD} = -2$ to find the equation of CD using the point-slope form $y - y_1 = m(x - x_1)$ .

1.1.3

Point B is the intersection of line AB and line CD.

Substitute equation (2) into equation (1) to solve for $x$ .

Substitute $x = -5$ into equation (2) to solve for $y$ .

1.1.4

Identify the center of the circle.

Calculate the radius $r$ .

Square the radius to get $r^2$ .

Substitute the center coordinates and $r^2$ into the standard equation of a circle.

Still stuck on this one?

More Mathematics Questions

In the diagram below, a circle with the centre at A (−3; −1) and a tangent CD to the circle at point B. Given also that CD cuts the x-axis at −6 and that the equation of the line segment AB (radius) is −x + 2y − 1 = 0.

Step-by-step solution

Question 1.1

1.1.1

Identify the given information.

Recall the geometric property that a radius drawn to the point of tangency is perpendicular to the tangent.

1.1.2

Find the gradient of the line segment AB.

Determine the gradient of the tangent CD.

Use the point C (−6,0)(-6, 0)(−6,0) (where CD cuts the xxx-axis at −6-6−6) and the gradient mCD=−2m_{CD} = -2mCD​=−2 to find the equation of CD using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1​=m(x−x1​).

1.1.3

Point B is the intersection of line AB and line CD.

Substitute equation (2) into equation (1) to solve for xxx.

Substitute x=−5x = -5x=−5 into equation (2) to solve for yyy.

1.1.4

Identify the center of the circle.

Calculate the radius rrr.

Square the radius to get r2r^2r2.

Substitute the center coordinates and r2r^2r2 into the standard equation of a circle.

Still stuck on this one?

More Mathematics Questions

Use the point C $(-6, 0)$ (where CD cuts the $x$ -axis at $-6$ ) and the gradient $m_{CD} = -2$ to find the equation of CD using the point-slope form $y - y_1 = m(x - x_1)$ .

Substitute equation (2) into equation (1) to solve for $x$ .

Substitute $x = -5$ into equation (2) to solve for $y$ .

Calculate the radius $r$ .

Square the radius to get $r^2$ .

Substitute the center coordinates and $r^2$ into the standard equation of a circle.