Tangent Circle Formula
A tangent of a circle in geometry is defined as a straight line that touches the circle at only one point. A tangent never enters the circle’s interior.
The tangent has two important properties:
 A tangent touches a circle at exactly one point on it.
 The tangent touches the circle’s radius at a right angle.
What is Tangent Circle Formula?
Let us now learn about the equation of the tangent. The tangent circle formula refers to the equation of the tangent and is written as:
\((y−y_0)=m_{tgt}(x−x_0)\)
Let us now have a look at a few solved examples using the tangent circle formula.
Solved Examples on Tangent Circle Formula

Example 1: Point \((1,5)\) lies on a curve given by \(y=f(x)=x^3x+5\). Find the equation of the tangent line to the curve that passes through the given point.
Solution: To write the equation of a line we need two things:
1. Slope
2. A point on the line
It is given that the curve contains a point \((1,5)\)
The slope is the same as the slope of the curve at \(x=1\) which is equal to the function’s derivative at that point:
\[\begin{align*} f(x)&=x^3x+5\\ f'(x)&=3x^2  1\\f'(x)&=3(1)^2  1=2\end{align*}\]
Substituting the slope \(m\) in the pointslope form of the line we would have:
\[\begin{align*} yy_0 &= m_{tangent}(xx_0)\\ y5 &=2(x1)\\ \end{align*}\]
Converting the above equation into \(y\)intercept form as:
\[\begin{align*} y5 &=2(x1)\\ y5&=2x2\\ y&=2x+3\end{align*}\]
Answer: \(\therefore\) The equation for the tangent line is \(y=2x13\) 
Example 2: Point \((1,5)\) lies on a curve given by \(y=f(x)=x^3x+5\). Find the equation of the tangent line to the curve that passes through the given point.
Solution:
To write the equation of a line we need two things:
1. Slope
2. A point on the line
It is given that the curve contains a point \((1,5)\)
The slope is the same as the slope of the curve at \(x=1\) which is equal to the function’s derivative at that point:
\[\begin{align*} f(x)&=x^3x+5\\ f'(x)&=3x^2  1\\f'(x)&=3(1)^2  1=2\end{align*}\]
Substituting the slope \(m\) in the pointslope form of the line we would have:
\[\begin{align*} yy_0 &= m_{tangent}(xx_0)\\ y5 &=2(x1)\\ \end{align*}\]
Converting the above equation into \(y\)intercept form as:
\[\begin{align*} y5 &=2(x1)\\ y5&=2x2\\ y&=2x+3\end{align*}\]
Answer: \(\therefore\) The equation for the tangent line is \(y=2x+3\)