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Chapter 1: Problem 10

The centre of a circle which is orthogonal to the circles \(C_{1}: x^{2}+y^{2}=25\) and \(C_{2}:(x-7)^{2}+(y-5)^{2}=4\) and which has the least radius is : (a) \(\left(5, \frac{25}{7}\right)\) (b) \((5,7)\) (c) \((7,12)\) (d) none of these

Short Answer

Expert verified
The answer is (a). The center of the circle which is orthogonal to the given circles and has the least radius is \(\left(5, \frac{25}{7}\right)\).

Step by step solution

01

Understand the Properties

Orthogonal circles are circles that intersect at right angles. Therefore, the radius of each circle is perpendicular to the radius of the other at the point of intersection.
02

Find the Centers

The center of \(C_1\) is at the origin (0,0) since the equation is in the form \(x^2 + y^2 = r^2\). For circle \(C_2\), the equation is in the form \((x - h)^2 + (y - k)^2 = r^2\), thus the center is (7,5).
03

Find the Line of Centers

The line connecting the two centers should have the equation \(y - 0 = \frac{5 - 0}{7 - 0}(x - 0)\), or simply \(y = \frac{5}{7}x\).
04

Find the Center of the Unknown Circle

Since the center of the unknown circle should be on the line of centers in order to minimize the radius, we can substitute \(y = \frac{5}{7}x\) into the equation of \(C_2\) to find the x-coordinate. This gives us \((x - 7)^2 + (\frac{5}{7}x - 5)^2 = 4\), which simplifies to \(x = 5\). Substituting \(x = 5\) into \(y = \frac{5}{7}x\) then gives us \(y = \frac{25}{7}\). Therefore, the center of the unknown circle is \(\left(5, \frac{25}{7}\right)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orthogonal Circles
Orthogonal circles are a fascinating concept in coordinate geometry. They are circles that intersect each other at right angles. This means that at the point where they cross, the radius lines of the two circles form a 90-degree angle. Understanding this relationship is crucial in solving problems involving such circles.
  • When dealing with orthogonal circles, each circle's radius is perpendicular to every radius of the other circle at the points they intersect.
  • This property makes orthogonal circles useful in defining unique geometrical relationships and constraints.
  • When two circles are orthogonal, the product of their slopes at the intersection points is -1, indicating perpendicular radii.
This perpendicularity is what defines the circles as orthogonal, which is a key property used in solving problems involving the intersection and construction of such circles.
Circle Equations
Circle equations are fundamental in coordinate geometry for representing these perfectly round shapes. The general equation of a circle is expressed as \[(x-h)^2 + (y-k)^2 = r^2\]where:
  • \(h, k\) are the coordinates of the center of the circle.
  • \(r\) is the radius of the circle.
For example, the circle equation \(x^2 + y^2 = 25\) represents a circle centered at the origin with a radius of 5, as 25 is the square of the radius. In another circle equation \((x-7)^2 + (y-5)^2 = 4\), we see the circle is centered at \((7,5)\) with a radius of 2.
  • This standard form of a circle equation helps us quickly identify the circle鈥檚 center and its radius, which are essential for further geometric analysis.
  • Understanding this form aids in the intersection and manipulation of circle properties, especially when applying additional constraints, like orthogonality.
Intersection of Circles
The intersection of circles occurs when two circles share one or two points. In the context of orthogonal circles, they're interesting because these points have perpendicular radii.
Finding where two circles intersect involves solving their equations simultaneously.
  • When two circles intersect, there's potential for them to have common tangents and intersect at exactly two points, one point (tangent), or none if they are separate.
  • For example, given the circle \(x^2 + y^2 = 25\) and \((x-7)^2 + (y-5)^2 = 4\), one can determine where these intersect by substituting one equation into the other or by rearranging to find solutions for \(x\) and \(y\).
Solving these intersections is crucial for tasks like finding an unknown circle that is orthogonal to the given circles and sharing a defined relationship with them. The understanding of intersection points aid in determining other geometric properties, such as orthogonality and relative positioning.

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Most popular questions from this chapter

If \(x_{1}, x_{2}, x_{3}\) as well as \(y_{1}, y_{2}, y_{3}\) are in GP with same common ratio, then the points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\) and \(\left(x_{3}, y_{3}\right)\) are : \(\begin{array}{ll}\text { (a) lie on a straight line } & \text { (b) lie on an ellipse }\end{array}\) (c) lie on a circle (d) are vertices of a triangle

Prove that the equation of the circumcircle of the triangle formed by the lines $$ \begin{aligned} &u_{1}=a_{1} x+b_{1} y+c_{1}=0 \\ &u_{2}=a_{2} x+b_{2} y+c_{2}=0 \\ &u_{3}=a_{3} x+b_{2} y+c_{1}=0 \end{aligned} $$ $$ \text { is }\left|\begin{array}{ccc} \frac{1}{u_{1}} & \frac{1}{u_{2}} & \frac{1}{u_{3}} \\ a_{2} a_{3}-b_{2} b_{3} & a_{3} a_{1}-b_{3} b_{1} & a_{1} a_{2}-b_{1} b_{2} \\\ a_{2} b_{3}+a_{3} b_{2} & a_{3} b_{1}+a_{1} b_{3} & a_{1} b_{2}+a_{2} b_{1} \end{array}\right|=0 $$ $$ \text { or }\left|\begin{array}{ccc} \frac{a_{1}^{2}+b_{1}^{2}}{u_{1}} & \frac{a_{2}^{2}+b_{2}^{2}}{u_{2}} & \frac{a_{3}^{2}+b_{3}^{2}}{u_{3}} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|=0 $$

AP and \(B Q\) are fixed parallel tangents to a circle and a tangent at any point \(C\) cuts them at \(P\) and \(Q\) respectively. Show that \(C P, C Q\) is independent of the position of \(C\) on the circle and \(\angle P O R\) is a right angie.

The medians \(A D\) and \(B E\) of the triangle with vertices \(A(0, b), B(0,0)\) and \(C(a, 0)\) are mutually perpendicular if : (a) \(b=a \sqrt{2}\) (b) \(a=b \sqrt{2}\) (c) \(b=-a \sqrt{2}\) (d) \(a=-b \sqrt{2}\)

Let the base of a triangle lie along the line \(x=a\) and be of length \(2 a .\) The area of this triangle is \(a^{2}\), if the vertex lies on the line :. (a) \(x=-a\) (b) \(x=0\) (c) \(x=\frac{a}{2}\) (d) \(x=2 a\)
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