Through a fixed point (h, k) secants...

MathematicsStandard and General Equation of a CircleJEE Advanced 1983Moderate

Through a fixed point $(h, k)$ secants are drawn to the circle $x^{2} + y^{2} = r^{2}$ . Show that the locus of the mid-points of the secants intercepted by the circle is $x^{2} + y^{2} = h x + k y$ .

Visualized Solution (Hindi)

Visualizing the Circle and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P (h, k)

Consider the circle $x^{2} + y^{2} = r^{2}$ with center at the origin $(0, 0)$ .
Let $P (h, k)$ be a fixed point in the plane.
A secant is a line passing through $P$ that intersects the circle at two distinct points.

Defining the Midpoint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M (x_{1}, y_{1})

Let $M (x_{1}, y_{1})$ be the midpoint of the chord formed by the secant.
Our goal is to find the relationship between $x_{1}$ and $y_{1}$ that holds for all such secants.

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T = S_{1} Equation

The equation of a chord of the circle $S = 0$ with a given midpoint $M (x_{1}, y_{1})$ is given by $T = S_{1}$ .
For the circle $x^{2} + y^{2} - r^{2} = 0$ :
$T = x x_{1} + y y_{1} - r^{2}$
$S_{1} = x_{1}^{2} + y_{1}^{2} - r^{2}$

Simplifying the Chord Equation

Equating $T$ and $S_{1}$ :
$x x_{1} + y y_{1} - r^{2} = x_{1}^{2} + y_{1}^{2} - r^{2}$
Canceling $- r^{2}$ from both sides:
$x x_{1} + y y_{1} = x_{1}^{2} + y_{1}^{2}$

Applying the Fixed Point Constraint

The secant passes through the fixed point $P (h, k)$ .
Therefore, the coordinates $(h, k)$ must satisfy the equation of the chord.

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (h, k) into the Equation

Substitute $x = h$ and $y = k$ into $x x_{1} + y y_{1} = x_{1}^{2} + y_{1}^{2}$ :
$h x_{1} + k y_{1} = x_{1}^{2} + y_{1}^{2}$
This is the condition satisfied by the midpoint $M (x_{1}, y_{1})$ .

Final Locus: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x^{2} + y^{2} = h x + k y

To find the locus, replace $(x_{1}, y_{1})$ with the general coordinates $(x, y)$ :
$h x + k y = x^{2} + y^{2}$
Rearranging the terms:
$x^{2} + y^{2} = h x + k y$
This represents a circle passing through the origin and the point $(h, k)$ .

00:00 / 00:00

Conceptually Similar Problems

MathematicsCondition for OrthogonalityModerate

View in:English Hindi

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = k^{2}$ orthogonally, then the equation of the locus of its centre is

(A)

2 a x + 2 b y - (a^{2} + b^{2} + k^{2}) = 0

(B)

2 a x + 2 b y - (a^{2} - b^{2} + k^{2}) = 0

(C)

x^{2} + y^{2} - 3 a x - 4 b y + (a^{2} + b^{2} - k^{2}) = 0

(D)

x^{2} + y^{2} - 2 a x - 3 b y + (a^{2} - b^{2} - k^{2}) = 0

Answer:A

MathematicsStandard and General Equation of a CircleJEE Advanced 1987Moderate

View in:English Hindi

Let a given line $L_{1}$ intersects the $x$ and $y$ axes at $P$ and $Q$ , respectively. Let another line $L_{2}$ , perpendicular to $L_{1}$ , cut the $x$ and $y$ axes at $R$ and $S$ , respectively. Show that the locus of the point of intersection of the lines $P S$ and $QR$ is a circle passing through the origin.

MathematicsCondition for OrthogonalityJEE Main 2005Moderate

View in:English Hindi

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = p^{2}$ orthogonally, then the equation of the locus of its centre is

(A)

x^{2} + y^{2} - 3 a x - 4 b y + (a^{2} + b^{2} - p^{2}) = 0

(B)

2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0

(C)

x^{2} + y^{2} - 2 a x - 3 b y + (a^{2} - b^{2} - p^{2}) = 0

(D)

2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0

Answer:D

MathematicsEquation of Tangent and NormalJEE Advanced 2016Moderate

View in:English Hindi

Let $R S$ be the diameter of the circle $x^{2} + y^{2} = 1$ , where $S$ is the point $(1, 0)$ . Let $P$ be a variable point (other than $R$ and $S$ ) on the circle and tangents to the circle at $S$ and $P$ meet at the point $Q$ . The normal to the circle at $P$ intersects a line drawn through $Q$ parallel to $R S$ at point $E$ . Then the locus of $E$ passes through the point(s)

* Multiple Correct Options

(A)

(\frac{1}{3}, \frac{1}{3})

(B)

(\frac{1}{4}, \frac{1}{2})

(C)

(\frac{1}{3}, - \frac{1}{3})

(D)

(\frac{1}{4}, - \frac{1}{2})

Answer:A, C

MathematicsCondition for OrthogonalityJEE Main 2004Moderate

View in:English Hindi

If a circle passes through the point $(a, b)$ and cuts the circle $x^{2} + y^{2} = 4$ orthogonally, then the locus of its centre is

(A)

2 a x - 2 b y - (a^{2} + b^{2} + 4) = 0

(B)

2 a x + 2 b y - (a^{2} + b^{2} + 4) = 0

(C)

2 a x - 2 b y + (a^{2} + b^{2} + 4) = 0

(D)

2 a x + 2 b y + (a^{2} + b^{2} + 4) = 0

Answer:B

MathematicsEquation of Tangent and NormalJEE Advanced 1985Moderate

View in:English Hindi

From the origin chords are drawn to the circle $(x - 1)^{2} + y^{2} = 1$ . The equation of the locus of the mid-points of these chords is .........

Answer:x^2 + y^2 - x = 0

MathematicsStandard and General Equation of a CircleModerate

View in:English Hindi

The locus of the mid-point of a chord of the circle $x^{2} + y^{2} = 4$ which subtends a right angle at the origin is

(A)

x + y = 2

(B)

x^{2} + y^{2} = 1

(C)

x^{2} + y^{2} = 2

(D)

x + y = 1

Answer:C

MathematicsEquation of Tangent and NormalJEE Advanced 2004Moderate

View in:English Hindi

If tangents are drawn to the ellipse $x^{2} + 2 y^{2} = 2$ , then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

(A)

\frac{1}{2 x ^{2}} + \frac{1}{4 y ^{2}} = 1

(B)

\frac{1}{4 x ^{2}} + \frac{1}{2 y ^{2}} = 1

(C)

\frac{x ^{2}}{2} + \frac{y ^{2}}{4} = 1

(D)

\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1

Answer:A

MathematicsStandard and General Equation of a CircleJEE Main 2006Moderate

View in:English Hindi

Let $C$ be the circle with centre $(0, 0)$ and radius 3 units. The equation of the locus of the mid points of the chords of the circle $C$ that subtend an angle of $\frac{2 π}{3}$ at its center is

(A)

x^{2} + y^{2} = \frac{3}{2}

(B)

x^{2} + y^{2} = 1

(C)

x^{2} + y^{2} = \frac{27}{4}

(D)

x^{2} + y^{2} = \frac{9}{4}

Answer:D

MathematicsStandard Equations of Parabola, Ellipse, and HyperbolaJEE Advanced 2001Moderate

View in:English Hindi

Let $P$ be a point on the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, 0 < b < a$ . Let the line parallel to y-axis passing through $P$ meet the circle $x^{2} + y^{2} = a^{2}$ at the point $Q$ such that $P$ and $Q$ are on the same side of x-axis. For two positive real numbers $r$ and $s$ , find the locus of the point $R$ on $P Q$ such that $P R : R Q = r : s$ as $P$ varies over the ellipse.

Answer:

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2} ( r + s ) ^{2}}{( b s + a r ) ^{2}} = 1