Animated Solution for Mathematics - Circles: Let the parabola $y=x^2+px-3$, meet the coordinate axes at the points P, Q and R. If the circle C with centre at $(-1,-1)$ passes through the points P, Q and R, then the area of $△PQR$ is:

Imagine you are standing on a coordinate plane, watching two geometric shapes perform a delicate dance. We have a parabola, $y=x^2+px-3$, sweeping across the grid, and a circle, centered at $(-1,-1)$, holding three specific points in its embrace: $P,Q,$ and $R$.

It feels like a complex puzzle, but as we peel back the layers, you will see that this problem is not about brute force; it is about finding the hidden symmetry.

Let us begin with the parabola defined as $y=x^2+px-3$. We are told it intersects the coordinate axes at points $P,Q,$ and $R$.

The most accessible point is $R$, the $y$-intercept. To find where any curve crosses the $y$-axis, we set $x=0$.

When we do this, the equation simplifies beautifully:

Thus, our point $R$ is fixed at $(0,-3)$. This is our first anchor in the coordinate plane.

Now, let us turn our attention to the circle. We know its center is at $(-1,-1)$.

The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$, where the center is $(-g,-f)$. By comparing this to our given center $(-1,-1)$, we identify that $g=1$ and $f=1$.

Our circle equation begins to take shape:

We are missing only one piece of the puzzle: the constant $c$. We know the circle passes through point $R(0,-3)$. Substituting these coordinates into our circle equation:

This simplifies to $9-6+c=0$, which leads us directly to $c=-3$. The mystery of the circle is solved: its equation is $x^2+y^2+2x+2y-3=0$.

Here is where the magic happens. The points $P$ and $Q$ are the $x$-intercepts of the parabola, but they also lie on our circle.

To find these points, we set $y=0$ in our circle equation. The equation becomes:

This is a classic quadratic equation. We can factorize it as:

The roots are $x=-3$ and $x=1$. Therefore, our points $P$ and $Q$ are $(-3,0)$ and $(1,0)$. Notice how we bypassed the parameter $p$ entirely; the geometry dictated the result.

We have arrived at the final act with our three vertices: $P(-3,0)$, $Q(1,0)$, and $R(0,-3)$.

The base of our triangle, the segment $PQ$, lies perfectly on the $x$-axis. The length of this base is the distance between $x=-3$ and $x=1$:

The height of the triangle is the perpendicular distance from $R(0,-3)$ to the $x$-axis, which is simply $|-3|=3$ units. The area of a triangle is given by the formula:

Substituting our values, we get:

The area of $△PQR$ is exactly $6$ square units. It is a beautiful, clean result, born from the elegant intersection of a parabola and a circle.