The equation of the circle withcentre at (4,3) and touching the line `5x-12y-10=0` is

Options

A. `x^(2)+y^(2)-4x-6y+4=0`
B. `x^(2)+y^(2)+6x-8y+16=0`
C. `x^(2)+y^(2)-8x-6y+21=0`
D. `x^(2)+y^(2)-24x-10y+144=0`

Answer

To find the equation of the circle with center at (4, 3) that touches the line given by the equation \(5x - 12y - 10 = 0\), we will follow these steps:

### Step 1: Identify the center of the circle
The center of the circle is given as \(C(4, 3)\).

### Step 2: Find the distance from the center to the line
To find the radius of the circle, we need to calculate the perpendicular distance from the center of the circle to the line. The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For the line \(5x - 12y - 10 = 0\), we have:
- \(A = 5\)
- \(B = -12\)
- \(C = -10\)

Substituting the center coordinates \((x_0, y_0) = (4, 3)\):

\[ d = \frac{|5(4) - 12(3) - 10|}{\sqrt{5^2 + (-12)^2}} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ 5(4) - 12(3) - 10 = 20 - 36 - 10 = -26 \]

Taking the absolute value:

\[ | -26 | = 26 \]

### Step 4: Calculate the denominator
Calculating the denominator:

\[ \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

### Step 5: Calculate the distance (radius)
Now substituting back into the distance formula:

\[ d = \frac{26}{13} = 2 \]

Thus, the radius \(r\) of the circle is \(2\).

### Step 6: Write the equation of the circle
The standard form of the equation of a circle with center \((h, k)\) and radius \(r\) is:

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

Substituting \(h = 4\), \(k = 3\), and \(r = 2\):

\[ (x - 4)^2 + (y - 3)^2 = 2^2 \]

This simplifies to:

\[ (x - 4)^2 + (y - 3)^2 = 4 \]

### Step 7: Expand the equation
Expanding the equation:

\[ (x^2 - 8x + 16) + (y^2 - 6y + 9) = 4 \]

Combining like terms:

\[ x^2 + y^2 - 8x - 6y + 25 - 4 = 0 \]

This simplifies to:

\[ x^2 + y^2 - 8x - 6y + 21 = 0 \]

### Final Result
The equation of the circle is:

\[ x^2 + y^2 - 8x - 6y + 21 = 0 \]

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