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Zorluk: Çok zorEquations of Circles in the Coordinate Plane

In the xyxy-plane, the graph of the equation x2+y2−14x−12y+q=0x^2 + y^2 - 14x - 12y + q = 0 is a circle that is tangent to the yy-axis, where qq is a constant. What is the value of qq?

Cevap: 36

Cevap

36
To find the value of qq, we convert the general form of the circle's equation into standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2 by completing the square. Grouping the variables gives (x2−14x)+(y2−12y)=−q(x^2 - 14x) + (y^2 - 12y) = -q. Adding (14/2)2=49(14/2)^2 = 49 and (12/2)2=36(12/2)^2 = 36 to both sides yields (x−7)2+(y−6)2=85−q(x - 7)^2 + (y - 6)^2 = 85 - q. Thus, the center of the circle is (7,6)(7, 6) and the radius squared is r2=85−qr^2 = 85 - q. Since the circle is tangent to the yy-axis, its radius must equal the horizontal distance from its center to the yy-axis, which is 77. Therefore, the radius squared is 72=497^2 = 49. Setting 85−q=4985 - q = 49 and solving for qq gives q=36q = 36.

Adım Adım Çözüm

1
Rearrange and group the terms of the equation to prepare for completing the square.
(x2−14x)+(y2−12y)=−q(x^2 - 14x) + (y^2 - 12y) = -q
Grouping terms allows us to complete the square for the xx and yy variables independently.
2
Complete the square for both the xx and yy expressions by adding the square of half the coefficient of the linear term to both sides.
(x2−14x+49)+(y2−12y+36)=−q+49+36(x^2 - 14x + 49) + (y^2 - 12y + 36) = -q + 49 + 36
Adding (14/2)2=49(14/2)^2 = 49 and (12/2)2=36(12/2)^2 = 36 transforms the trinomials into perfect squares.
3
Factor the perfect square trinomials and simplify the right side of the equation.
(x−7)2+(y−6)2=85−q(x - 7)^2 + (y - 6)^2 = 85 - q
This puts the equation in the standard form of a circle, (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, identifying the center as (7,6)(7, 6) and the radius squared as r2=85−qr^2 = 85 - q.
4
Determine the radius of the circle using the condition of tangency to the yy-axis.
r=7r = 7
A circle tangent to the yy-axis has a radius equal to the absolute value of the x-coordinate of its center, which is ∣7∣=7|7| = 7.
5
Solve for the constant qq by equating the two expressions for the radius squared.
85−q=49  ⟹  q=3685 - q = 49 \implies q = 36
Since r=7r = 7, we have r2=49r^2 = 49. Setting 85−q=4985 - q = 49 yields q=36q = 36.

Anahtar Kavram

Completing the square to find the standard equation of a circle and applying coordinate geometry tangency conditions.
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Konular

Equations of Circles in the Coordinate Plane

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