Answer :
To prove that the straight line [tex]\( x = 2y + 5 \)[/tex] is tangent to the circle [tex]\( x^2 + y^2 = 5 \)[/tex] algebraically, we need to show that the line intersects the circle at exactly one point.
1. Substitute the equation of the line into the equation of the circle:
The equation of the line is:
[tex]\[ x = 2y + 5 \][/tex]
Substitute [tex]\( x \)[/tex] from the line equation into the circle equation:
[tex]\[ (2y + 5)^2 + y^2 = 5 \][/tex]
2. Expand and simplify:
[tex]\[ (2y + 5)^2 + y^2 = 5 \][/tex]
Expanding [tex]\( (2y + 5)^2 \)[/tex]:
[tex]\[ 4y^2 + 20y + 25 \][/tex]
So, the equation becomes:
[tex]\[ 4y^2 + 20y + 25 + y^2 = 5 \][/tex]
Combine like terms:
[tex]\[ 5y^2 + 20y + 25 = 5 \][/tex]
Subtract 5 from both sides to set it to 0:
[tex]\[ 5y^2 + 20y + 20 = 0 \][/tex]
3. Solve the quadratic equation:
The general form of the quadratic equation is [tex]\( ay^2 + by + c = 0 \)[/tex]. Here, [tex]\( a = 5 \)[/tex], [tex]\( b = 20 \)[/tex], and [tex]\( c = 20 \)[/tex].
The discriminant of a quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Compute the discriminant:
[tex]\[ \Delta = 20^2 - 4 \cdot 5 \cdot 20 \][/tex]
[tex]\[ \Delta = 400 - 400 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
4. Conclusion from the discriminant:
When the discriminant ([tex]\( \Delta \)[/tex]) is 0, the quadratic equation has exactly one real solution. This implies that the line [tex]\( x = 2y + 5 \)[/tex] intersects the circle [tex]\( x^2 + y^2 = 5 \)[/tex] at precisely one point. Therefore, the line is tangent to the circle.
This completes the algebraic proof that the straight line [tex]\( x = 2y + 5 \)[/tex] is tangent to the circle [tex]\( x^2 + y^2 = 5 \)[/tex].
1. Substitute the equation of the line into the equation of the circle:
The equation of the line is:
[tex]\[ x = 2y + 5 \][/tex]
Substitute [tex]\( x \)[/tex] from the line equation into the circle equation:
[tex]\[ (2y + 5)^2 + y^2 = 5 \][/tex]
2. Expand and simplify:
[tex]\[ (2y + 5)^2 + y^2 = 5 \][/tex]
Expanding [tex]\( (2y + 5)^2 \)[/tex]:
[tex]\[ 4y^2 + 20y + 25 \][/tex]
So, the equation becomes:
[tex]\[ 4y^2 + 20y + 25 + y^2 = 5 \][/tex]
Combine like terms:
[tex]\[ 5y^2 + 20y + 25 = 5 \][/tex]
Subtract 5 from both sides to set it to 0:
[tex]\[ 5y^2 + 20y + 20 = 0 \][/tex]
3. Solve the quadratic equation:
The general form of the quadratic equation is [tex]\( ay^2 + by + c = 0 \)[/tex]. Here, [tex]\( a = 5 \)[/tex], [tex]\( b = 20 \)[/tex], and [tex]\( c = 20 \)[/tex].
The discriminant of a quadratic equation [tex]\( ay^2 + by + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Compute the discriminant:
[tex]\[ \Delta = 20^2 - 4 \cdot 5 \cdot 20 \][/tex]
[tex]\[ \Delta = 400 - 400 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
4. Conclusion from the discriminant:
When the discriminant ([tex]\( \Delta \)[/tex]) is 0, the quadratic equation has exactly one real solution. This implies that the line [tex]\( x = 2y + 5 \)[/tex] intersects the circle [tex]\( x^2 + y^2 = 5 \)[/tex] at precisely one point. Therefore, the line is tangent to the circle.
This completes the algebraic proof that the straight line [tex]\( x = 2y + 5 \)[/tex] is tangent to the circle [tex]\( x^2 + y^2 = 5 \)[/tex].
