Question Statement

A normal line cuts the circle with centre at $(1,2)$ at the point $(3,5)$. Find the other point of intersection of the normal to the circle. Also, find the equation of the circle.

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Background and Explanation

A normal to a circle is a line perpendicular to the tangent at the point of contact. A key geometric property is that every normal line of a circle passes through its center. Consequently, the segment of the normal line between the two points where it intersects the circle forms a diameter, meaning the center is the midpoint of these two points.

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Solution

Part (a): Finding the other point of intersection

Let $A(3,5)$ be the given point of intersection and $B(x,y)$ be the other point of intersection.

Since the normal line always passes through the centre of the circle, the centre $C(1,2)$ is the midpoint of the diameter $AB$. Using the midpoint formula:

$\left(\frac{x+3}{2},\frac{y+5}{2}\right)=(1,2)$

We can solve for $x$ and $y$ by equating the coordinates:

For the x-coordinate: $\frac{x+3}{2}=1$ $x+3=2$ $x=-1$

For the y-coordinate: $\frac{y+5}{2}=2$ $y+5=4$ $y=-1$

Thus, the other point of intersection of the normal with the circle is $B(-1,-1)$.

Part (b): Finding the equation of the circle

To find the equation of the circle, we first need to determine the radius $r$. The radius is the distance from the centre $C(1,2)$ to any point on the circle, such as $A(3,5)$.

Using the distance formula: $r=|AC|=\sqrt{(3-1{)}^2+(5-2{)}^2}$ $r=\sqrt{2^2+3^2}$ $r=\sqrt{4+9}$ $r=\sqrt{13}$

Now, we use the standard form of the circle equation $(x-h{)}^2+(y-k{)}^2=r^2$, where $(h,k)$ is the centre $(1,2)$:

$(x-1{)}^2+(y-2{)}^2=(\sqrt{13}{)}^2$

Expanding the squares: $x^2-2x+1+y^2-4y+4=13$

Combining like terms and setting the equation to zero: $x^2+y^2-2x-4y+5=13$ $x^2+y^2-2x-4y-8=0$

The equation of the circle is $x^2+y^2-2x-4y-8=0$.

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Key Formulas or Methods Used

• Property of Normals: Every normal to a circle passes through the center.
• Midpoint Formula: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
• Standard Equation of a Circle: $(x-h{)}^2+(y-k{)}^2=r^2$

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Summary of Steps

1. Identify that the center $(1,2)$ is the midpoint between the two intersection points of the normal.
2. Use the midpoint formula to calculate the coordinates of the second point $B(-1,-1)$.
3. Calculate the radius of the circle by finding the distance between the center and point $A(3,5)$.
4. Substitute the center and the squared radius into the circle equation formula.
5. Expand and simplify the equation into general form.