Locus problem?

Question

Show that circles x^2+y^2=4 and x^2+2x+y^2-4y-4=0 both have 3x+4y+10=0 as a tangent. 

Please show all working out

Answer 1

If two circles have same tangent, then the distances from the center to the tangent line is equal to the radius of the circle.

Distance from the line ax + by + c = 0 to the point (x₀, y₀) is | ax₀ + by₀ + c|/√(a² + b² )

 

The circles are : x² + y² = 4 → (1)

                     x² + 2x + y² - 4y -  4 = 0 → (2).

The standard form of the circle equation is ( x - h )² + ( y - k )² = r², where, (h, k) is the center of the circle, and r is the radius.

Compare equation (1) with standard form of the circle equation.

Center = (0, 0) and radius = 2.

The tangent line is 3x + 4y + 10 = 0 and the center is (0, 0).

The distance from 3x + 4y + 10 = 0 to (0, 0) = radius of the circle

| [(3*0) + (4*0) + 10] |/√(3² + 4² )

= | 0 + 0 + 10 |/√(9 + 16)

= | 10 |/√25

= 10/5

= 2

= radius of the circle.

Write equation (2) :  x² + 2x + y² - 4y -  4 = 0 in standard form of a circle.

To change the expression into a perfect square  add (half the x coefficient)² and add (half the y coefficient)²to each side of the expression.

Here x coefficient = 2, so, (half the x coefficient)² = (2/2)²= 1.

Here y coefficient = - 4, so, (half the y coefficient)² = (- 4/2)²= 4.

Add 1 and 4 to each side.

x² + 2x + 1 + y² - 4y + 4 - 4 = 1 + 4

(x + 1)² + (y - 2)² = 4 + 5

(x + 1)² + (y - 2)² = 9

(x - (- 1))² + (y - 2)² = 3²

Compare the equation with standard form of a circle equation.

Center = (- 1, 2) and radius = 3.

The tangent line is 3x + 4y + 10 = 0 and the center is (- 1, 2).

The distance from 3x + 4y + 10 = 0 to (- 1, 2) = radius of the circle

| [(3*- 1) + (4*2) + 10] |/√(3² + 4² )

= | - 3 + 8 + 10 |/√(9 + 16)

= | 15 |/√25

= 15/5

= 3

= radius of the circle.

Hence proved that, the circles x² + y² = 4 and x² + 2x + y² - 4y -  4 = 0 both have 3x + 4y + 10 = 0 as the tangent.