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Sagot :
To find the equation of the tangent to the circle given by the equation [tex]\( x^2 + y^2 + 4x - 6y - 13 = 0 \)[/tex] at the point [tex]\( (3, 4) \)[/tex], follow these steps:
1. Rewrite the Circle Equation in Standard Form:
The general form of a circle's equation is [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]. We need to rewrite the given equation in this form.
[tex]\[ x^2 + 4x + y^2 - 6y - 13 = 0 \][/tex]
Complete the square for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \begin{aligned} x^2 + 4x &\implies (x + 2)^2 - 4, \\ y^2 - 6y &\implies (y - 3)^2 - 9. \end{aligned} \][/tex]
Thus, the equation becomes:
[tex]\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 - 13 = 0. \][/tex]
Simplify:
[tex]\[ (x + 2)^2 + (y - 3)^2 - 26 = 0, \][/tex]
[tex]\[ (x + 2)^2 + (y - 3)^2 = 26. \][/tex]
This represents a circle centered at [tex]\( (-2, 3) \)[/tex] with a radius of [tex]\( \sqrt{26} \)[/tex].
2. Find the Gradient of the Tangent Line:
The gradient of the tangent line at a point on a circle is perpendicular to the radius at that point. The gradient of the radius (from the center to the point [tex]\( (3, 4) \)[/tex]) can be found as follows:
The center of the circle is [tex]\( (-2, 3) \)[/tex], and the point of tangency is [tex]\( (3, 4) \)[/tex].
The gradient (slope) of the radius [tex]\( m_{\text{radius}} \)[/tex] is:
[tex]\[ m_{\text{radius}} = \frac{4 - 3}{3 - (-2)} = \frac{1}{5}. \][/tex]
Since the tangent is perpendicular to the radius, the gradient [tex]\( m_{\text{tangent}} \)[/tex] is:
[tex]\[ m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -5. \][/tex]
3. Write the Equation of the Tangent Line:
Using the point-slope form of the line equation [tex]\( y - y_1 = m(x - x_1) \)[/tex] with [tex]\( m = -5 \)[/tex] and the point [tex]\( (3, 4) \)[/tex]:
[tex]\[ y - 4 = -5(x - 3), \][/tex]
[tex]\[ y - 4 = -5x + 15, \][/tex]
[tex]\[ y = -5x + 19. \][/tex]
To write the equation in standard form [tex]\( Ax + By + C = 0 \)[/tex]:
[tex]\[ 5x + y - 19 = 0. \][/tex]
Simplify:
[tex]\[ 5x + y = 19. \][/tex]
4. Match with the Given Options:
We see that this matches the option (c):
c. [tex]\( 5x + y = 19 \)[/tex].
So, the correct answer is:
c. [tex]\( 5x + y = 19 \)[/tex].
1. Rewrite the Circle Equation in Standard Form:
The general form of a circle's equation is [tex]\( (x - h)^2 + (y - k)^2 = r^2 \)[/tex]. We need to rewrite the given equation in this form.
[tex]\[ x^2 + 4x + y^2 - 6y - 13 = 0 \][/tex]
Complete the square for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ \begin{aligned} x^2 + 4x &\implies (x + 2)^2 - 4, \\ y^2 - 6y &\implies (y - 3)^2 - 9. \end{aligned} \][/tex]
Thus, the equation becomes:
[tex]\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 - 13 = 0. \][/tex]
Simplify:
[tex]\[ (x + 2)^2 + (y - 3)^2 - 26 = 0, \][/tex]
[tex]\[ (x + 2)^2 + (y - 3)^2 = 26. \][/tex]
This represents a circle centered at [tex]\( (-2, 3) \)[/tex] with a radius of [tex]\( \sqrt{26} \)[/tex].
2. Find the Gradient of the Tangent Line:
The gradient of the tangent line at a point on a circle is perpendicular to the radius at that point. The gradient of the radius (from the center to the point [tex]\( (3, 4) \)[/tex]) can be found as follows:
The center of the circle is [tex]\( (-2, 3) \)[/tex], and the point of tangency is [tex]\( (3, 4) \)[/tex].
The gradient (slope) of the radius [tex]\( m_{\text{radius}} \)[/tex] is:
[tex]\[ m_{\text{radius}} = \frac{4 - 3}{3 - (-2)} = \frac{1}{5}. \][/tex]
Since the tangent is perpendicular to the radius, the gradient [tex]\( m_{\text{tangent}} \)[/tex] is:
[tex]\[ m_{\text{tangent}} = -\frac{1}{m_{\text{radius}}} = -5. \][/tex]
3. Write the Equation of the Tangent Line:
Using the point-slope form of the line equation [tex]\( y - y_1 = m(x - x_1) \)[/tex] with [tex]\( m = -5 \)[/tex] and the point [tex]\( (3, 4) \)[/tex]:
[tex]\[ y - 4 = -5(x - 3), \][/tex]
[tex]\[ y - 4 = -5x + 15, \][/tex]
[tex]\[ y = -5x + 19. \][/tex]
To write the equation in standard form [tex]\( Ax + By + C = 0 \)[/tex]:
[tex]\[ 5x + y - 19 = 0. \][/tex]
Simplify:
[tex]\[ 5x + y = 19. \][/tex]
4. Match with the Given Options:
We see that this matches the option (c):
c. [tex]\( 5x + y = 19 \)[/tex].
So, the correct answer is:
c. [tex]\( 5x + y = 19 \)[/tex].
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