Problem 106
Question
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
Step-by-Step Solution
Verified Answer
The assertion and reason correctly imply that \(25 < k < 32\).
1Step 1: 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 \((3, 5)\).
2Step 2: 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 \(f = -5\).
3Step 3: 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: \[-32 + k < 0\].
4Step 4: Solve for k using Interior Point Criterion
From \(-32 + k < 0\), we have \(k < 32\). Thus, for the point to be interior, \(k < 32\).
5Step 5: 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 \(25 < k\).
6Step 6: Conclude the Inequality for k
Combine both conditions: \(25 < k < 32\).
Key Concepts
Circle EquationInterior PointInequality for k
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:
- \(x^2 + y^2 - 6x - 10y + k = 0\)
- \((x - h)^2 + (y - k)^2 = r^2\),
- is found at \((3, 5)\).
- This occurs through rearranging terms and observing the coefficients of \(x\) and \(y\).
- The center is \((h,k) = (3, 5)\),
- and its radius \(r\)
Interior Point
Determining whether a point lies inside a circle involves utilizing the circle's equation. A given point
- such as \((2, 4)\)
- \(x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c < 0\)
- is inside the circle.
- \(x_1 = 2\) and
- \(y_1 = 4\),
- \(x^2 + y^2 - 6x - 10y + k\),
- \(-32 + k < 0\).
- if the point is internal or not based on the value of \(k\).
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
- \((2, 4)\)
- to be inside, \(-32 + k < 0\), and
- therefore, \(k < 32\).
- 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 \(k > 25\).
- \(25 < k < 32\)
Other exercises in this chapter
Problem 103
Assertion: The locus of the centres of circles passing through the origin and cutting the circle \(x^{2}+y^{2}+6 x-\) \(4 y+2=0\) orthogonally is \(3 x-2 y+1=0\
View solution Problem 104
Assertion: The tangent to the circle \(x^{2}+y^{2}=5\) at the point \((1,-2)\) also touches the circle \(x^{2}+y^{2}-8 x+6 y+\) \(20=0\). Then its point of cont
View solution Problem 107
Assertion: The equation of the circle passing through the point \((2 a, 0)\) and whose radical axis is \(x=\frac{a}{2}\) with respect to the circle \(x^{2}+y^{2
View solution Problem 110
The greatest distance of the point \(P(10,7)\) from the circle \(x^{2}+y^{2}-4 x-2 y-20=0\) is [2002| (A) 10 unit (B) 15 unit (C) 5 unit (D) none of these
View solution