Problem 6
Question
If the equations of four circles are \((x \pm 4)^{2}+(y \pm 4)^{2}\) \(=4^{2}\), then the radius of the smallest circle touching all the four circles is (A) \(4(\sqrt{2}+1)\) (B) \(4(\sqrt{2}-1)\) (C) \(2(\sqrt{2}-1)\) (D) none of these
Step-by-Step Solution
Verified Answer
The radius of the smallest circle is (B) \(4(\sqrt{2}-1)\).
1Step 1: Understand the positions of the circles
The given equations represent four circles centered at \((4, 4), (-4, 4), (4, -4), (-4, -4)\) all with radius 4. Visualizing the positions, these centers form a square of side 8 centered at the origin (0,0).
2Step 2: Determine the distance to the new circle's center
To find the smallest circle touching all four circles, consider the center at the origin with radius extending to the nearest point of each circle. Distance from the origin (0,0) to any given center like (4,4) is calculated by \[\text{distance} = \sqrt{4^2 + 4^2} = 4\sqrt{2}\].
3Step 3: Calculate the required radius
The circle we want must reach a distance just enough to intersect the other circles' radii, meaning the new circle's radius is the calculated distance minus the radius of the given circles: \[\text{radius} = 4\sqrt{2} - 4\]. Simplifying gives the smallest radius as \[4(\sqrt{2} - 1)\].
4Step 4: Match the answer to the options
From the simplified radius, the choice that matches is (B) \(4(\sqrt{2}-1)\).
Key Concepts
Radius CalculationSquare GeometryCoordinates of a Circle Center
Radius Calculation
Understanding how to calculate the radius of a circle is a fundamental skill in circle geometry. In this exercise, we are introduced to the concept by examining circles with the standard form of the circle equation: \[ (x - h)^2 + (y - k)^2 = r^2 \] where
- \((h, k)\) is the center of the circle
- \(r\) is the radius.
Square Geometry
Square geometry finds its application in identifying and analyzing patterns when dealing with multiple circles’ placements. Here, the circle centers
Identifying this square helps us quantify relationships, like where a new circle’s center might be. Frequently, square geometry ensures that we can utilize right-angled triangle properties and the Pythagorean theorem to successfully locate other important measurements or centers.
- \((4, 4), (-4, 4), (4, -4), (-4, -4)\)
Identifying this square helps us quantify relationships, like where a new circle’s center might be. Frequently, square geometry ensures that we can utilize right-angled triangle properties and the Pythagorean theorem to successfully locate other important measurements or centers.
Coordinates of a Circle Center
Locating the circle centers involves interpreting each given equation to find
To solve the problem, the new circle we seek is also centered at \((0, 0)\). This uniform distribution aids in making decisions about extending or adjusting radius values based on calculated distances. Each step requires a keen interpretation of how coordinates affect the circle's placement in relation with others—a valuable insight for geometrical problem-solving.
- \((h, k)\), the coords of the centers and
- how all relate to each other spatially.
To solve the problem, the new circle we seek is also centered at \((0, 0)\). This uniform distribution aids in making decisions about extending or adjusting radius values based on calculated distances. Each step requires a keen interpretation of how coordinates affect the circle's placement in relation with others—a valuable insight for geometrical problem-solving.
Other exercises in this chapter
Problem 4
If the line \((y-2)=m(x+1)\) intersects the circle \(x^{2}+\) \(y^{2}+2 x-4 y-3=0\) at two real distinct points, then the number of possible values of \(m\) is
View solution Problem 5
The number of points on the circle \(x^{2}+y^{2}-4 x-10 y+\) \(13=0\) which are at a distance 1 from the point \((-3,2)\) is (A) \(\underline{1}\) (B) 2 (C) 3 (
View solution Problem 7
The intercept on the line \(y=x\) by the circle \(x^{2}+y^{2}-2 x=\) 0 is \(A B\). Equation of the circle with \(A B\) as a diameter is (A) \(x^{2}+y^{2}+x+y=0\
View solution Problem 8
The locus of the mid-point of the chord of the circle \(x^{2}\) \(+y^{2}-2 x-2 y-2=0\) which makes an angle of \(120^{\circ}\) at the centre is (A) \(x^{2}+y^{2
View solution