Question

The points of intersection of the line y=x+2 and the circle (x-2)² + y² = 16 are

• A. (-2,0), (2,4)
• B. (-2,4), (2,0)
• C. (4,0), (4,2)
• D. (4,6), (4,-2)
• E. (4,0), (4,-2)

Answer 1

The Correct Option is A

We are given the equation of a line and a circle:
Line: \( y = x + 2 \) - Circle: \( (x - 2)^2 + y^2 = 16 \)
To find the points of intersection, we substitute the equation of the line into the equation of the circle.
Step 1: Substitute \( y = x + 2 \) into the circle equation Substitute \( y = x + 2 \) into the circle equation: \[ (x - 2)^2 + (x + 2)^2 = 16 \]
Step 2: Expand both terms First, expand \( (x - 2)^2 \): \[ (x - 2)^2 = x^2 - 4x + 4 \] Now expand \( (x + 2)^2 \): \[ (x + 2)^2 = x^2 + 4x + 4 \] Substitute both expansions into the equation: \[ x^2 - 4x + 4 + x^2 + 4x + 4 = 16 \] Simplify: \[ 2x^2 + 8 = 16 \]
Step 3: Solve for \( x \) Now subtract 8 from both sides: \[ 2x^2 = 8 \] Divide by 2: \[ x^2 = 4 \] Take the square root of both sides: \[ x = \pm 2 \] ### Step 4: Find the corresponding \( y \) values For \( x = 2 \), substitute into the line equation \( y = x + 2 \): \[ y = 2 + 2 = 4 \] For \( x = -2 \), substitute into the line equation \( y = x + 2 \): \[ y = -2 + 2 = 0 \]

The correct option is (A) : \((-2,0), (2,4)\)

Answer 2

We are given the line \(y = x + 2\) and the circle \((x - 2)^2 + y^2 = 16\).

To find the points of intersection, we substitute \(y = x + 2\) into the equation of the circle:

\((x - 2)^2 + (x + 2)^2 = 16\)

Expanding the terms, we get:

\(x^2 - 4x + 4 + x^2 + 4x + 4 = 16\)

Combining like terms, we have:

\(2x^2 + 8 = 16\)

Subtracting 8 from both sides, we get:

\(2x^2 = 8\)

Dividing by 2, we have:

\(x^2 = 4\)

Taking the square root, we get:

\(x = \pm 2\)

When \(x = 2\), \(y = x + 2 = 2 + 2 = 4\). So, one point of intersection is \((2, 4)\).

When \(x = -2\), \(y = x + 2 = -2 + 2 = 0\). So, the other point of intersection is \((-2, 0)\).

Therefore, the points of intersection are \((-2, 0)\) and \((2, 4)\).