Dimensional Geometry - 8 - Free MCQ Practice Test with solutions, Maths

The given equation of the circle is
x² + y² - 2hx - 2ky + h² = 0
or (x-h)² + (y-k)² - k²= 0
or (x - h)² + (y - k)² = k²
Thus the radius = y- coordinate of the centre and therefore circle touch the axis of x.

The given equation of the circle is

The general equation

The general equation of the circle
x² + y² + 2gx + 2fy + c = 0
involves three unknowns, namely, g, f and c. Therefore exactly three conditions are required for determining the circle and no more.
Remark: It follows that a unique circle will always pass through three non-collinear points in the plane.

Proof: Let the circle be described on the line segment joining A(x₁, y₁) and B(x₂, y₂) as diameter. Consider a point P(x, y) on the circle (other than A and B). Then we know from geometry that

This is the required equation.
Comments: Centre C of this circle is the middle point of AB

Let the circle cut the axis of x at the points A and R. Then
y = 0 for these points.
Substituting y = 0 in the given equation of the circle, we get

Remark: Note that in obtaining the equation of the tangent to the circle at a given point (V ,i j ) , we replace in the equation of the circle as follows:

We are given the following:

Their points of intersection will be obtained by- solving equations (i) and (ii). Eliminating y from (i) and (ii), we get 

The roots of this equation and hence the points of intersection of (i) and (ii) are real only if 

Equation of a line passing through a point (x₁, y₁, z₁) and having d.c.’s [l,m,n].
Let P(x, y, z) be any point on the line.
Then the d.c.’s of the line joining A and P are


This is th required equation of the lline.

Proof: the d.c.’s of the two given lines are [3, 2k, 2] and (3k, 1,-5) Since the lines are perpendicular, therefore

The d.c.’s of the line perpendicular to x-axis are (0, m, n)
∴ Required equation will be

The equations of the line in the symmetrical form are

where (x₁, y₁ ,z₁) is a given point on the line and (l, m, n) are the direction ratios of the line. Therefore the equations of a straight line can be written in symmetrical form if we know.
(i) the direction ratios of the line and
(ii) a point on it.

 We know that any two equations of first dgree in x, y and z taken together represent a line [two non parallel planes always intersect in a line].
The equations of the line are

i.e.  the line is the intersection of planes (i) and (ii) Let (l, m, n) be the d.c.'s of the line of intersection of planes (i) and (ii). Since the line is common to both the planes, therefore it is perpendicular to the normals to the two planes. But the direction ratios of these normals are the coefficients of x, y and x in the equations of the planes. Therefore we have

Let (l₁, m₁, n₁) be the d.c.’s of the line of intersection of the planes

To find the angle between the line


Let θ be 1 he angle between the line (i) and the plane (ii). Then 90 - θ is the angle between the line and the normal to the plane.
Now th2 d.r.’s of the line (i) are l, m, n and the d.r/s of the line perpendicular lo (ii) are a, b, c
∴ cos (90 - θ)