Question

The intercept made by the circle ${x^2} + {y^2} - 14x - 10y + 23 = 0$ on the line \[7x + y - 4 = 0\] is
A) 1
B) 2
C) 3
D)4

Answer 1

Hint- In order to find the intercept made by the given circle and the line first we have to find the radius and center of the circle by comparing the given equation with the general equation of circle, then we will proceed further by using the distance formula from a centre to the given line.

Complete step-by-step answer:
Given Circle equation is ${x^2} + {y^2} - 14x - 10y + 23 = 0$
 And line equation is \[7x + y - 4 = 0\]

From the given figure
AB is the intercept made by circle and line
So we have to find the length of AB
First we will find the radius of the given circle
We know that
The General Form of the equation of a circle is \[{x^2} + {y^2} + 2gx + 2fy + c = 0\] and the centre of the circle is \[\left( { - g, - f} \right)\] and the radius is \[\sqrt {\left( {{g^2} + {f^2} - c} \right)} \]
By comparing the given circle equation to the general circle equation we have
Centre as \[\left( {7,5} \right)\]
And the radius will be
Radius $ = \sqrt {\left( {{7^2} + {5^2} - 23} \right)} = \sqrt {\left( {49 + 25 - 23} \right)} = \sqrt {51} $
In right angle triangle AOM, we have OA as a radius, and to find AM, we have to find OM
So, as we know that distance from the point \[\left( {{x_1},{y_1}} \right)\] on the line \[ax + by + c = 0\] is given as
$D = \dfrac{{\left| {a{x_1} + b{y_1} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
Therefore the distance from the centre \[\left( {7,5} \right)\] to the line will be
$
  D = \dfrac{{\left| {7\left( 7 \right) + \left( 5 \right) + \left( { - 4} \right)} \right|}}{{\sqrt {{7^2} + {1^2}} }} \\
   \Rightarrow D = \dfrac{{50}}{{\sqrt {50} }} = \sqrt {50} \\
 $
$\therefore OM = \sqrt {50} $
In right angle triangle AOM, we will apply Pythagoras theorem in order to find AM
$
   \Rightarrow {\left( {AO} \right)^2} = {\left( {MO} \right)^2} + {\left( {AM} \right)^2} \\
   \Rightarrow {\left( {AM} \right)^2} = {\left( {AO} \right)^2} - {\left( {MO} \right)^2} \\
   \Rightarrow AM = \sqrt {{{\left( {AO} \right)}^2} - {{\left( {MO} \right)}^2}} \\
   \Rightarrow AM = \sqrt {51 - 50} \\
   \Rightarrow AM = 1 \\
 $
And we know that AB = 2AM
So intercept AB is given as:
$ \Rightarrow AB = 2 \times 1 = 2$
Hence the value of intercept made by the line on the circle is 2
So, the correct answer is option B.

Note- The intercepts of a line are the points where the line intercepts, or crosses, the horizontal and vertical axes. The straight line on the graph below intercepts the two coordinate axes. The point where the line crosses the x-axis is called the [x-intercept]. In order to solve such problems students must remember the distance formula in coordinate geometry for different conditions.