Question Statement

A circle has its centre at the point $C (0, 1)$ and a line touches the circle. The point $P (3, 5)$ lies on the line touching the circle. The distance between $P$ and $C$ is five times the radius of the circle. Find the equation of the circle and the point where the line touches the circle.

Background and Explanation

To solve this problem, we use the standard equation of a circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ . We also apply the geometric property that the radius of a circle is perpendicular to the tangent line at the point of contact, meaning the product of their slopes is $- 1$ .

Solution

Let $A (a, b)$ be the point on the circle where the line touches the circle (the point of tangency) and $r$ be the radius of the circle.

Part 1: Finding the Radius and Equation of the Circle

We are given that the distance between $P (3, 5)$ and $C (0, 1)$ is five times the radius $r$ : $∣ P C ∣ = 5 r$

Using the distance formula: $(3 - 0)^{2} + (5 - 1)^{2} 3^{2} + 4^{2} 9 + 16 255 r = 5 r = 5 r = 5 r = 5 r = 5 r = 1$

Now, we can write the equation of the circle with centre $C (0, 1)$ and radius $r = 1$ : $(x - 0)^{2} + (y - 1)^{2} x^{2} + y^{2} - 2 y + 1 x^{2} + y^{2} - 2 y = 1^{2} = 1 = 0$

Part 2: Finding the Point of Tangency $A (a, b)$

Since the point $A (a, b)$ lies on the circle, it must satisfy the circle's equation: $a^{2} + b^{2} - 2 b = 0 \dots (i)$

The radius $\overline{A C}$ is perpendicular to the tangent line $\overline{A P}$ . Therefore, the product of their slopes is $- 1$ : $(slope of \overline{A C}) \times (slope of \overline{A P}) = - 1$

$(\frac{b - 1}{a - 0}) (\frac{b - 5}{a - 3}) \frac{b ^{2} - 5 b - b + 5}{a ^{2} - 3 a} b^{2} - 6 b + 5 b^{2} - 6 b + 5 a^{2} + b^{2} - 3 a - 6 b + 5 = - 1 = - 1 = - (a^{2} - 3 a) = - a^{2} + 3 a = 0 \dots (ii)$

To find $a$ and $b$ , we subtract equation (ii) from equation (i): $(a^{2} + b^{2} - 2 b) - (a^{2} + b^{2} - 3 a - 6 b + 5) 3 a + 4 b - 5 4 b b = 0 = 0 = 5 - 3 a = \frac{5 - 3 a}{4}$

Substitute $b = \frac{5 - 3 a}{4}$ into equation (i): $a^{2} + (\frac{5 - 3 a}{4})^{2} - 2 (\frac{5 - 3 a}{4}) a^{2} + \frac{25 - 30 a + 9 a ^{2}}{16} - \frac{5 - 3 a}{2} = 0 = 0$

Multiply the entire equation by 16 to clear the denominators: $16 a^{2} + (25 - 30 a + 9 a^{2}) - 8 (5 - 3 a) 16 a^{2} + 25 - 30 a + 9 a^{2} - 40 + 24 a 25 a^{2} - 6 a - 15 = 0 = 0 = 0$

Using the quadratic formula $a = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ : $a a = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( 25 ) ( - 15 )}{2 ( 25 )} = \frac{6 \pm 36 + 1500}{50} = \frac{6 \pm 1536}{50} = \frac{6 \pm 256 \times 6}{50} = \frac{6 \pm 16 6}{50} = \frac{3 \pm 8 6}{25}$

Now, find the corresponding values for $b$ :

Case 1: $a = \frac{3 + 8 6}{25}$ $b b = \frac{1}{4} [5 - 3 (\frac{3 + 8 6}{25})] = \frac{1}{4} [\frac{125 - 9 - 24 6}{25}] = \frac{1}{4} [\frac{116 - 24 6}{25}] = \frac{29 - 6 6}{25}$

Case 2: $a = \frac{3 - 8 6}{25}$ $b b = \frac{1}{4} [5 - 3 (\frac{3 - 8 6}{25})] = \frac{1}{4} [\frac{125 - 9 + 24 6}{25}] = \frac{1}{4} [\frac{116 + 24 6}{25}] = \frac{29 + 6 6}{25}$

The points of tangency are: $(\frac{3 + 8 6}{25}, \frac{29 - 6 6}{25}) and (\frac{3 - 8 6}{25}, \frac{29 + 6 6}{25})$

Key Formulas or Methods Used

Distance Formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
Circle Equation: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Perpendicular Slopes: $m_{1} \cdot m_{2} = - 1$
Slope Formula: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Summary of Steps

Calculate the distance $∣ P C ∣$ using the coordinates of the center and the given point.
Use the relation $∣ P C ∣ = 5 r$ to solve for the radius ( $r = 1$ ).
Write the equation of the circle using the center $(0, 1)$ and $r = 1$ .
Set up two equations for the point of tangency $A (a, b)$ : one from the circle's equation and one from the perpendicularity of the radius and tangent.
Solve the resulting system of equations to find the coordinates of the point $A$ .

Part 1: Finding the Radius and Equation of the Circle

We are given that the distance between

P (3, 5)

and

C (0, 1)

is five times the radius

r

∣ P C ∣ = 5 r

Using the distance formula:

(3 - 0)^{2} + (5 - 1)^{2} 3^{2} + 4^{2} 9 + 16 255 r = 5 r = 5 r = 5 r = 5 r = 5 r = 1

Now, we can write the equation of the circle with centre

C (0, 1)

and radius

r = 1

(x - 0)^{2} + (y - 1)^{2} x^{2} + y^{2} - 2 y + 1 x^{2} + y^{2} - 2 y = 1^{2} = 1 = 0

Part 2: Finding the Point of Tangency

A (a, b)

Since the point

A (a, b)

lies on the circle, it must satisfy the circle's equation:

a^{2} + b^{2} - 2 b = 0 \dots (i)

The radius

\overline{A C}

is perpendicular to the tangent line

\overline{A P}

. Therefore, the product of their slopes is

- 1

(slope of \overline{A C}) \times (slope of \overline{A P}) = - 1

(\frac{b - 1}{a - 0}) (\frac{b - 5}{a - 3}) \frac{b ^{2} - 5 b - b + 5}{a ^{2} - 3 a} b^{2} - 6 b + 5 b^{2} - 6 b + 5 a^{2} + b^{2} - 3 a - 6 b + 5 = - 1 = - 1 = - (a^{2} - 3 a) = - a^{2} + 3 a = 0 \dots (ii)

To find

a

and

b

, we subtract equation (ii) from equation (i):

(a^{2} + b^{2} - 2 b) - (a^{2} + b^{2} - 3 a - 6 b + 5) 3 a + 4 b - 5 4 b b = 0 = 0 = 5 - 3 a = \frac{5 - 3 a}{4}

Substitute

b = \frac{5 - 3 a}{4}

into equation (i):

a^{2} + (\frac{5 - 3 a}{4})^{2} - 2 (\frac{5 - 3 a}{4}) a^{2} + \frac{25 - 30 a + 9 a ^{2}}{16} - \frac{5 - 3 a}{2} = 0 = 0

Multiply the entire equation by 16 to clear the denominators:

16 a^{2} + (25 - 30 a + 9 a^{2}) - 8 (5 - 3 a) 16 a^{2} + 25 - 30 a + 9 a^{2} - 40 + 24 a 25 a^{2} - 6 a - 15 = 0 = 0 = 0

Using the quadratic formula

a = \frac{- b \pm b ^{2} - 4 a c}{2 a}

a a = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( 25 ) ( - 15 )}{2 ( 25 )} = \frac{6 \pm 36 + 1500}{50} = \frac{6 \pm 1536}{50} = \frac{6 \pm 256 \times 6}{50} = \frac{6 \pm 16 6}{50} = \frac{3 \pm 8 6}{25}

Now, find the corresponding values for

b

Case 1: $a = \frac{3 + 8 6}{25}$

b b = \frac{1}{4} [5 - 3 (\frac{3 + 8 6}{25})] = \frac{1}{4} [\frac{125 - 9 - 24 6}{25}] = \frac{1}{4} [\frac{116 - 24 6}{25}] = \frac{29 - 6 6}{25}

Case 2: $a = \frac{3 - 8 6}{25}$

b b = \frac{1}{4} [5 - 3 (\frac{3 - 8 6}{25})] = \frac{1}{4} [\frac{125 - 9 + 24 6}{25}] = \frac{1}{4} [\frac{116 + 24 6}{25}] = \frac{29 + 6 6}{25}

The points of tangency are:

(\frac{3 + 8 6}{25}, \frac{29 - 6 6}{25}) and (\frac{3 - 8 6}{25}, \frac{29 + 6 6}{25})

Summary of Steps

Calculate the distance

∣ P C ∣

using the coordinates of the center and the given point.

Use the relation

∣ P C ∣ = 5 r

to solve for the radius (

r = 1

Write the equation of the circle using the center

(0, 1)

and

r = 1

Set up two equations for the point of tangency

A (a, b)

: one from the circle's equation and one from the perpendicularity of the radius and tangent.

Solve the resulting system of equations to find the coordinates of the point

A

7.1 Q-13

Question Statement

Background and Explanation

Solution

Part 1: Finding the Radius and Equation of the Circle

Part 2: Finding the Point of Tangency $A (a, b)$

Key Formulas or Methods Used

Summary of Steps

7.1 Q-13

Question Statement

Background and Explanation

Solution

Part 1: Finding the Radius and Equation of the Circle

Part 2: Finding the Point of Tangency $A (a, b)$

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Part 1: Finding the Radius and Equation of the Circle

Part 2: Finding the Point of Tangency A(a,b)

Key Formulas or Methods Used

Summary of Steps

Question Statement

Background and Explanation

Solution

Part 1: Finding the Radius and Equation of the Circle

Part 2: Finding the Point of Tangency A(a,b)

Key Formulas or Methods Used

Summary of Steps

Part 2: Finding the Point of Tangency $A (a, b)$

Part 2: Finding the Point of Tangency $A (a, b)$