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Chapter 19: Problem 106

Assertion: If the point \((2,4)\) is interior to the circle \(x^{2}\) \(+y^{2}-6 x-10 y+k=0\) and the circle does not cut the axes at any point, then \(25

Short Answer

Expert verified
The assertion and reason correctly imply that \(25 < k < 32\).

Step by step solution

01

Understand the Circle Equation

The given equation of the circle is \( x^2 + y^2 - 6x - 10y + k = 0 \). This can be rewritten in standard circle form \((x - h)^2 + (y - k)^2 = r^2\) after completing the square. Thus, compare this equation with \((x - 3)^2 + (y - 5)^2 = r^2\). Here, the circle has center \((3, 5)\).
02

Determine the Criterion for Interior Point

According to the Reason, if a point \((x_1, y_1)\), lies inside the circle given by \(x^2 + y^2 + 2gx + 2fy + c = 0\), then \(x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c < 0\). For the circle \( x^2 + y^2 - 6x - 10y + k = 0 \), we identify \(2g = -6\), \(2f = -10\), so \(g = -3\) and \(f = -5\).
03

Substitute Point into the Inequality

Substitute \((x_1, y_1) = (2, 4)\) into the inequality: \[2^2 + 4^2 - 6 \times 2 - 10 \times 4 + k < 0\]. Compute: \[4 + 16 - 12 - 40 + k < 0\]. Simplify to get: \[-32 + k < 0\].
04

Solve for k using Interior Point Criterion

From \(-32 + k < 0\), we have \(k < 32\). Thus, for the point to be interior, \(k < 32\).
05

Use the Condition Circle Does Not Cut Axes

The circle does not cut the x-axis and y-axis. The circle equation implies if it does not cut the axes, then \(c > 0\) is required, where \(c = k\). Thus, \(k > 25\) because the radius is less than 0 when equal to 25. Evaluate where \((3, 0)\text{ to not cut x-axis and }(0, 5)\text{ to not cut y-axis} \), these give lower bounds with \(25 < k\).
06

Conclude the Inequality for k

Combine both conditions: \(25 < k < 32\).

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Key Concepts

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

Circle Equation
In coordinate geometry, the equation of a circle plays a crucial role in understanding the properties and positions of circles on the Cartesian plane. The general form of a circle's equation is given by:
  • \(x^2 + y^2 - 6x - 10y + k = 0\)
To convert this into the standard form, which is
  • \((x - h)^2 + (y - k)^2 = r^2\),
requires completing the square. For our specific example, the circle's center
  • is found at \((3, 5)\).
  • This occurs through rearranging terms and observing the coefficients of \(x\) and \(y\).
The circle鈥檚 equation tells us key features:
  • The center is \((h,k) = (3, 5)\),
  • and its radius \(r\)
can be calculated once \(k\) is known. This understanding helps us analyze if points lie within this circle.
Interior Point
Determining whether a point lies inside a circle involves utilizing the circle's equation. A given point
  • such as \((2, 4)\)
can be plugged into the equation of the circle to verify its position. The general rule is:
  • \(x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c < 0\)
if the point
  • is inside the circle.
In our problem, by substituting
  • \(x_1 = 2\) and
  • \(y_1 = 4\),
into
  • \(x^2 + y^2 - 6x - 10y + k\),
we find
  • \(-32 + k < 0\).
This inequality helps us measure
  • if the point is internal or not based on the value of \(k\).
When this inequality holds correct, the point is confirmed to lie within the circle defined by our formula.
Inequality for k
The variable \(k\) in the circle equation impacts the positioning and size of the circle on the plane. To determine acceptable values for \(k\), ensure both the point
  • \((2, 4)\)
is inside the circle and that the circle doesn't intersect the coordinate axes. For the point
  • to be inside, \(-32 + k < 0\), and
  • therefore, \(k < 32\).
To ensure the circle doesn't cut the x or y-axis:
  • note that \(c\), or equivalently \(k\) in the original equation, must be positive.
  • The radius squared must be larger than zero due to \((3\),
  • 0) and \((0\), involving the formula
  • 5).
  • This gives a lower bound such that \(k > 25\).
By combining these constraints, the inequality
  • \(25 < k < 32\)
ensures both conditions are satisfied, securely placing \(k\) in a range where the circle fits all given conditions.

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Most popular questions from this chapter

If \(\theta_{1}, \theta_{2}\) be the inclinations of tangents drawn from the point \(\vec{P}\) to the circle \(x^{2}+y^{2}=a^{2}\) and \(\cot \theta_{1}+\cot \theta_{2}=k\), then the locus of \(P\) is (A) \(k\left(y^{2}+a^{2}\right)=2 x y\) (B) \(k\left(y^{2}-a^{2}\right)=2 x y\) (C) \(k\left(y^{2}-a^{2}\right)=x y\) (D) none of these

The pole of the chord of the circle \(x^{2}+y^{2}=16\) which is bisected at the point \((-2,3)\), with respect to the circle is (A) \(\left(\frac{-32}{13}, \frac{48}{13}\right)\) (B) \(\left(\frac{32}{13}, \frac{48}{13}\right)\) (C) \(\left(\frac{-32}{13}, \frac{-48}{13}\right)\) (D) none of these

The point on the straight line \(y=2 x+11\) which is nearest to the circle \(16\left(x^{2}+y^{2}\right)+32 x-8 y-50=0\) is (A) \(\left(\frac{9}{2}, 2\right)\) (B) \(\left(-\frac{9}{2}, 2\right)\) (C) \(\left(\frac{9}{2},-2\right)\) (D) none of these

If the chord of contact of tangents from a point on the circle \(x^{2}+y^{2}=a^{2}\) to the circle \(x^{2}+y^{2}=b^{2}\) touches the circle \(x^{2}+y^{2}=c^{2}\), then \(a, b, c\) are in (A) A. \(P\). (B) G. P. (C) H. P. (D) none of these

The point diametrically opposite to the point \(P(1,0)\) on the circle \(x^{2}+y^{2}+2 x+4 y-3=0\) is |2008| (A) \((3,-4)\) (B) \((-3,4)\) (C) \((-3,-4)\) (D) \((3,4)\)
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