A circle in the standard (x, y) coordinate plane is defined by the equation x^2 + y^2 - 8x… — ACT

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + y^2 - 8x - 6y = 0$ . A point on this circle has an $x$ -coordinate of $1$ and a positive $y$ -coordinate. What is the $y$ -coordinate of this point?

Answer: 7

Answer

The positive y-coordinate of the point on the circle is 7.

Substituting

x = 1

into the circle's equation

x^2 + y^2 - 8x - 6y = 0

gives

1 + y^2 - 8 - 6y = 0

, which simplifies to the quadratic equation

y^2 - 6y - 7 = 0

. Factoring this equation yields

(y - 7)(y + 1) = 0

. The solutions are

y = 7

and

y = -1

. Because the y-coordinate must be positive, the correct value is 7.

Step-by-Step Solution

Substitute the x-coordinate

x = 1

into the circle's equation.

1^2 + y^2 - 8(1) - 6y = 0

This sets up the equation to solve for the corresponding y-coordinates.

Simplify the equation.

y^2 - 6y - 7 = 0

Combining the constant terms creates a standard quadratic equation in terms of y.

Solve the quadratic equation by factoring.

(y - 7)(y + 1) = 0

, yielding

y = 7

y = -1

Factoring allows us to find the two possible y-values that satisfy the equation.

Choose the positive y-coordinate.

y = 7

The question specifies that the y-coordinate must be positive.

Key Concept

Substituting a known coordinate into a circle's equation to find the other coordinate using quadratic equations.

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