A Circle Having Radius 5 Units And Passing Through Point (4,3) And Its Centre Lies On Line 2x+Y-1=0.. Find The Equation Of Circle

A circle having radius 5 units and passing through point (4,3) and its centre lies on line 2x+y-1=0.. Find the equation of circle

Answer:

(x - 1)² + (y + 1)² = 5²

(x + 1)² + (y - 3)² = 5²

Step-by-step explanation:

We know that the equation of a circle is expressed as

(x - h)² + (y - k)² = r²

Given: r = 5

Unknown: (h, k)

To solve for the center, were given the following clues;

(1) the circle is passing through point (4, 3)

(2) the center lies on the line 2x + y - 1 = 0

Identify the two equations in which you can get the center of the circle;

Equation 1 = 2x + y - 1 = 0

Equation 2 = $\sqrt{(x-4)^{2}+(y-3)^{2} } =25$

*I used the distance formula in equation 2. This will help me identify the missing center of the circle.

Solve for y in equation 1:

2x + y - 1 = 0

2x + y = 1

y = 1 - 2x

Substitute the value of y in equation 2:

$\sqrt{(x-4)^{2}+(y-3)^{2} } =25$

$\sqrt{(x-4)^{2}+(1-2x-3)^{2} } =25$

$\sqrt{(x-4)^{2}+(-2x-2)^{2} } =25$

$x^{2} -8x+16+4x^{2} +8x+4=25\\5x^{2} +20=25\\5x^{2} -5=0\\x^{2} -1=0\\x=1\\x=-1$

Substitute the value of x in equation 1:

When x = 1, y = -1

When x = -1, y = 3

Substitute the values of x and y to (h, k). Yoll then get:

(x - 1)² + (y + 1)² = 5²

(x + 1)² + (y - 3)² = 5²

^_^

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