Question

Find the equation of the circle with centre (1, 2) and tangent \[x + y - 5 = 0\]
A.\[{x^2} + {y^2} + 2x - 4y + 6 = 0\]
B. \[{x^2} + {y^2} - 2x - 4y + 3 = 0\]
C. \[{x^2} + {y^2} - 2x - 4y - 8 = 0\]
D. \[{x^2} + {y^2} - 2x - 4y + 8 = 0\]

Answer 1

Hints First draw the diagram of the circle with a tangent line to have a clear concept. Use the formula of perpendicular distance from a line to a point to obtain the perpendicular distance from a line \[x + y - 5 = 0\] to a point (1, 2). The positive value of the distance is the radius of the circle. By using the standard form of circle, we will find the equation of circle.

Formula used
The formula of perpendicular distance of a line \[ax + by + c = 0\] to a point \[({x_1},{y_1})\]is
\[\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}\] .
The general equation of a circle with centre (h, k) and radius r is
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] .

Complete step by step solution
The diagram of the given problem is,

Image: Circle with tangent
Therefore, to obtain the radius of the given circle obtain the perpendicular distance from the given line to the centre.
So, the perpendicular distance is,
\[\dfrac{{1.1 + 1.2 - 5}}{{\sqrt {{1^2} + {1^2}} }}\]
\[ = - \dfrac{2}{{\sqrt 2 }}\]
\[ = - \sqrt 2 \] , but radius cannot be negative.
Therefore, the radius is \[\sqrt 2 .\]
We know the general equation of a circle with centre (h, k) and radius r is
\[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}\] .
So, the equation of the circle with radius (1, 2) and radius \[\sqrt 2 \] is,
\[{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} = {\left( {\sqrt 2 } \right)^2}\]
\[{x^2} - 2x + 1 + {y^2} - 4y + 4 = 2\]
\[{x^2} + {y^2} - 2x - 4y + 3 = 0\]
The correct option is B.

Note The diagram is the important part in this type of questions. Without the diagram it is not clear that the perpendicular distance from the centre to the tangent line is the radius of the circle. So, draw a diagram first to understand the question clearly.