Question

1. find the length of the tangent from the point (5,7) to the circle ( x^2 + y^2 - 4x - 6y + 9 = 0 )

Explanation:

Step1: Recall the formula for the length of the tangent from a point \((x_1,y_1)\) to the circle \(x^2 + y^2 + Dx + Ey + F = 0\)

The formula for the length of the tangent \(L\) from the point \((x_1,y_1)\) to the circle \(x^2 + y^2 + Dx + Ey + F = 0\) is given by \(L=\sqrt{x_1^2 + y_1^2 + Dx_1 + Ey_1 + F}\) (where the point \((x_1,y_1)\) lies outside the circle). First, we need to identify the values of \(D\), \(E\), \(F\) from the given circle equation and the coordinates of the point \((x_1,y_1)\).

Given the circle equation \(x^2 + y^2 - 4x - 6y + 9 = 0\), we can compare it with the general form \(x^2 + y^2 + Dx + Ey + F = 0\). So, \(D=- 4\), \(E = - 6\), \(F = 9\). The point from which we are drawing the tangent is \((x_1,y_1)=(5,7)\).

Step2: Substitute the values into the formula

Substitute \(x_1 = 5\), \(y_1=7\), \(D=-4\), \(E=-6\), \(F = 9\) into the formula \(L=\sqrt{x_1^2 + y_1^2 + Dx_1 + Ey_1 + F}\)

First, calculate \(x_1^2 + y_1^2 + Dx_1 + Ey_1 + F\):

\(x_1^2=5^2 = 25\), \(y_1^2=7^2=49\), \(Dx_1=-4\times5=-20\), \(Ey_1=-6\times7 = - 42\), \(F = 9\)

Now, sum these values:

\(25+49-20 - 42+9\)

\(=(25 + 49)+(9)-(20 + 42)\)

\(=74 + 9-62\)

\(=83 - 62\)

\(=21\)

Then, the length of the tangent \(L=\sqrt{21}\)

Answer:

The length of the tangent is \(\sqrt{21}\)