What is the distance from a point on the x axis to the center of the circle x2 plus y2 -2x -6y plus 5 equals 0 when the tangent of the circle is at 3 4?

If I understand your question you are asking for the distance from the point of intersection of the x-axis and the tangent to the circle at (3, 4) to the centre of the circle.

To solve this you need to:

1. Find the centre of the circle (X, Y) - a circle with equation (x - X)² + (y - Y)² = r² has a centre of (X, Y) and radius r;
2. Find the slope m of the radius from the centre of the circle and the point of contact of the tangent;
3. Use the slope m of the radius to calculate the slope m' of the tangent - the slopes of two perpendicular lines is such that mm' = -1;
4. Use the slope-point equation of a line to find the equation of the tangent - a line with slope m through point (X, Y) has equation: y - Y = m(x - X);
5. Find the point where this line crosses the x-axis
6. Use Pythagoras to find the distance from this point to the centre of the circle.

Have a go before reading the solution below

Hint: You need to complete the squares in x and y to rearrange the equation for the circle into the form in step 1.

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1. Find the centre of the circle x² + y² - 2x - 6y + 5 = 0

x² + y² - 2x - 6y + 5 = 0

→ x² - 2x + y² - 6y + 5 = 0

→ (x - 2/2)² - (2/2)² + (y - 6/2)² - (6/2)² + 5 = 0

→ (x - 1)² - 1 + (y - 3)² - 9 + 5 = 0

→ (x - 1)² + (y - 3)² = 5

→ centre of circle is at (1, 3)

2. Find the slope of the radius to (3, 4)

slope = change_in_y/change_in_x

→ m = (4 - 3)/ (3 - 1)

= 1/2

3. Calculate the slope of the tangent

mm' = -1

→ m' = -1/m

= -1(1/2)

= -2

→ slope of tangent is -2

4. Find the equation of the tangent

y - 4 = -2(x - 3)

→ y - 4 = -2x + 6

→ y + 2x = 10

5. Find the point where the tangent crosses the x-axis

x-axis is the line y = 0

→ y + 2x = 10

→ 0 + 2x = 10

→ 2x = 10

→ x = 5

→ tangent crosses x-axis at (5, 0)

6. Find the distance using Pythagoras

distance = √(difference_in_x² + difference_in_y²)

= √((5 - 1)² + (3 - 0)²)

= √(4² + 3²)

= √(16 + 9)

= √25

= 5

→ The distance from the intersection of the x-axis and the tangent to the circle x² + y² - 2x - 6y + 5 = 0 at (3, 4) to the centre of that circle is 5 units.