Question

Prove algebraically that the straight line with the equation x=2y+5 is a tangent to a circle with the equation x^2 + y^2 =5

Answer 1

Hi,

Explanation:

`\left\{\begin{array}{ccc}x&=&2y+5\\x^2+y^2&=&5\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}x&=&2y+5\\(2y+5)^2+y^2&=&5\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}x&=&2y+5\\5(y^2+4y+4)&=&0\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}x&=&2y+5\\4y^2+^20y+25+y^2&=&5\\\end {array} \right.\\\\\\\left\{\begin{array}{ccc}y&=&-2\\x&=&1\\\end {array} \right.\\\\\\`

The solution (1,-2) has a multiplicity 2.
The line is tangent to the circle.

To prove that the straight line with the equation x = 2y + 5 is a tangent to the circle with the equation x^2 + y^2 = 5, we need to show that the line intersects the circle at a single point. In other words, we need to find the point of intersection and demonstrate that the line only touches the circle at that point, making it a tangent.

Let's start by solving the system of equations:

x = 2y + 5

x^2 + y^2 = 5

Substitute the expression for x from the first equation (x = 2y + 5) into the second equation:

(2y + 5)^2 + y^2 = 5

Now, expand and simplify the equation:

4y^2 + 20y + 25 + y^2 = 5

Combine like terms:

5y^2 + 20y + 25 - 5 = 0

5y^2 + 20y + 20 = 0

Divide the entire equation by 5 to simplify further:

y^2 + 4y + 4 = 0

Now, we have a quadratic equation. To solve it, we can use the discriminant (b^2 - 4ac) to determine the nature of the roots:

a = 1

b = 4

c = 4

Discriminant (D) = b^2 - 4ac = 4^2 - 4(1)(4) = 16 - 16 = 0

Since the discriminant is zero, it means that the quadratic equation has a single real root. This means that the line intersects the circle at one point. Therefore, the line x = 2y + 5 is tangent to the circle x^2 + y^2 = 5 at that point.

So, algebraically, we have shown that the line is a tangent to the circle.