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Question: The equations to the circles which touch the lines \[3x - 4y + 1 = 0,\,4x + 3y - 7 = 0\] and pass th...

The equations to the circles which touch the lines 3x−4y+1=0, 4x+3y−7=03x - 4y + 1 = 0,\,4x + 3y - 7 = 0 and pass through (2,3)(2,3) are
A. x2+y2−4x−16y+43=0, 5x2+5y2−12x−24y+31=0{x^2} + {y^2} - 4x - 16y + 43 = 0,\,5{x^2} + 5{y^2} - 12x - 24y + 31 = 0
B. x2+y2+4x+16y−43=0, 5x2+5y2−12x−24y+31=0{x^2} + {y^2} + 4x + 16y - 43 = 0,\,5{x^2} + 5{y^2} - 12x - 24y + 31 = 0
C. x2+y2−4x−16y+43=0, 5x2+5y2+12x+24y+31=0{x^2} + {y^2} - 4x - 16y + 43 = 0,\,5{x^2} + 5{y^2} + 12x + 24y + 31 = 0
D. x2+y2+4x+16y−43=0, 5x2+5y2+12x+24y+31=0{x^2} + {y^2} + 4x + 16y - 43 = 0,\,5{x^2} + 5{y^2} + 12x + 24y + 31 = 0

Explanation

Solution

Hint : Use the distance formula for calculating the distance between the centre of the circle and the given lines and we are also given the passing point of the circle, this way we can find out the coordinates of the centre of the circle and its radius. Thus we can obtain the equation of the circle after identifying these unknown values.

Complete step-by-step answer :
The given lines touch the circle means that they are tangent to the circle at the point of contact. The line joining the centre of the circle and the point is the radius and is perpendicular to the tangent.
We know that the perpendicular distance of a line having equation ax+by+c=0ax + by + c = 0 from a point (x0,y0)({x_0},{y_0}) is given by the formula d=∣ax0+by0+c∣a2+b2d = \dfrac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2} + {b^2}} }}
Let coordinates of the centre be (h,k)(h,k) and its radius be rr .
The distance of centre from line 3x−4y+1=03x - 4y + 1 = 0 is, r=∣3h−4k+1∣5r = \dfrac{{\left| {3h - 4k + 1} \right|}}{5}
Squaring both sides, we get –
r2=(3h−4k+1)225{r^2} = \dfrac{{{{(3h - 4k + 1)}^2}}}{{25}}
The distance of centre from line 4x+3y−7=04x + 3y - 7 = 0 is, r=∣4h+3k−7∣5r = \dfrac{{\left| {4h + 3k - 7} \right|}}{5}
Squaring both sides, we get –
r2=(4h+3k−7)225{r^2} = \dfrac{{{{(4h + 3k - 7)}^2}}}{{25}}
Also, the circle passes through the point (2,3)(2,3) so, (h−2)2+(k−3)2=r2{(h - 2)^2} + {(k - 3)^2} = {r^2}
Comparing the above three equations and then solving them, we get h=65 or h=2h = \dfrac{6}{5}\,or\,h = 2 and k=125 or k=8k = \dfrac{{12}}{5}\,or\,k = 8
So, r=1 or r=5r = 1\,or\,r = 5
Putting these values in the general equation of the circle, we get –
⇒(x−65)2+(y−125)2=(1)2 ⇒x2+3625−12x5+y2+14425−24y5=1 ⇒x2+y2−12x5−24y5+365=1 ⇒5x2+5y2−12x−24y+36=5 ⇒5x2+5y2−12x−24y+31=0   \Rightarrow {(x - \dfrac{6}{5})^2} + {(y - \dfrac{{12}}{5})^2} = {(1)^2} \\\ \Rightarrow {x^2} + \dfrac{{36}}{{25}} - \dfrac{{12x}}{5} + {y^2} + \dfrac{{144}}{{25}} - \dfrac{{24y}}{5} = 1 \\\ \Rightarrow {x^2} + {y^2} - \dfrac{{12x}}{5} - \dfrac{{24y}}{5} + \dfrac{{36}}{5} = 1 \\\ \Rightarrow 5{x^2} + 5{y^2} - 12x - 24y + 36 = 5 \\\ \Rightarrow 5{x^2} + 5{y^2} - 12x - 24y + 31 = 0 \;
Or
⇒(x−2)2+(y−8)2=(5)2 ⇒x2+4−4x+y2+64−16y=25 ⇒x2+y2−4x−16y+43=0   \Rightarrow {(x - 2)^2} + {(y - 8)^2} = {(5)^2} \\\ \Rightarrow {x^2} + 4 - 4x + {y^2} + 64 - 16y = 25 \\\ \Rightarrow {x^2} + {y^2} - 4x - 16y + 43 = 0 \;
So, the correct answer is “Option A”.

Note : In a circle, all the points are equidistant from a single point called the centre of the circle. The distance between centre and any point on the circle is called the radius. The tangent at any point of the circle is perpendicular to the line joining the centre and the point; the distance of the given line from the centre of the given circle is calculated by the formula of finding the perpendicular distance of a point from a line.