Question

The equation of tangent to the circle (x-5)²+y² = 25 at (2, 4) is

• A. 3x-4y+10=0
• B. x+y=6
• C. 2x-y=0
• D. 3x-2y+2=0
• E. 3x-4y-10=0

Answer 1

The Correct Option is A

The equation of the circle is \( (x - 5)^2 + y^2 = 25 \). The center of the circle is \( (5, 0) \), and the radius is \( r = \sqrt{25} = 5 \). The point of tangency is \( (2, 4) \). The tangent at this point is perpendicular to the radius at this point.
Step 1: Find the slope of the radius The slope of the radius is given by the formula for the slope between two points: \[ \text{slope of the radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 0}{2 - 5} = \frac{4}{-3} = -\frac{4}{3} \]
Step 2: Find the slope of the tangent The slope of the tangent line is the negative reciprocal of the slope of the radius. Therefore, the slope of the tangent is: \[ \text{slope of the tangent} = \frac{3}{4} \]
Step 3: Use the point-slope form of the line equation The point-slope form of a line is given by: \[ y - y_1 = m(x - x_1) \] where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. Substituting \( m = \frac{3}{4} \) and the point \( (2, 4) \), we get: \[ y - 4 = \frac{3}{4}(x - 2) \] Simplifying: \[ y - 4 = \frac{3}{4}x - \frac{3}{2} \] \[ 4(y - 4) = 3x - 6 \] \[ 4y - 16 = 3x - 6 \] \[ 3x - 4y + 10 = 0 \]

Thus, the correct option is (A) : \(3x-4y+10=0\)

Answer 2

We are given the equation of the circle: \((x-5)^2 + y^2 = 25\) and the point (2, 4) on the circle. Our goal is to determine the equation of the tangent to the circle at this given point.

First, observe that the center of the circle is located at the point (5, 0), and the radius of the circle is 5.

Next, we calculate the slope of the radius that connects the center of the circle (5, 0) to the point (2, 4) on the circle. This slope, denoted as \(m_r\), is given by: \[m_r = \frac{4-0}{2-5} = \frac{4}{-3} = -\frac{4}{3}\]

Since the tangent line to the circle at the point (2, 4) is perpendicular to the radius at that point, the slope of the tangent line, denoted as \(m_t\), is the negative reciprocal of the slope of the radius. Thus, \[m_t = -\frac{1}{m_r} = -\frac{1}{-\frac{4}{3}} = \frac{3}{4}\]

Now that we have the slope of the tangent line, \(m_t = \frac{3}{4}\), and a point on the tangent line, (2, 4), we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by: \[y - y_1 = m_t(x - x_1)\] Substituting the values, we get: \[y - 4 = \frac{3}{4}(x - 2)\]

To simplify and express the equation in the standard form, we multiply both sides by 4: \[4(y - 4) = 3(x - 2)\] \[4y - 16 = 3x - 6\]

Finally, we rearrange the equation to the form \(Ax + By + C = 0\): \[3x - 4y + 10 = 0\]

Therefore, the equation of the tangent to the circle \((x-5)^2 + y^2 = 25\) at the point (2, 4) is determined to be \(3x - 4y + 10 = 0\).