Question

The equation of the circle passing through the point (1, –2) and having its centre on the line 2x – y – 14 = 0 and touching the line 4x + 3y – 23 = 0

• A. $x^2$ + $y^2$ + 8x + 12y + 27 = 0
• B. $x^2$ + $y^2$ - 12y + 27 = 0
• C. $x^2$ + $y^2$ - 8x + 12y - 27 = 0
• D. $x^2$ + $y^2$ - 8x + 12y + 27 = 0

Answer 1

The correct option is D $x^2$ + $y^2$ - 8x + 12y + 27 = 0

Let P(1, -2) and the centre be C(a, b)
Centre lies on 2x - y - 14 = 0 $\Rightarrow$ 2a - b - 14 = 0 $\Rightarrow$ b = 2a - 14 $\to$ (1)
Circle touches the lines 4x + 3y - 23 = 0. Lets say that this point of touching is Q.
We know that,
CQ = CP = Radius of the circle.
$\Rightarrow$ $\sqrt{{(a-1)}^2+{(b+2)}^2}$ = $\frac{|4a+3b-23|}{\sqrt{4^2+3^2}}$
$\Rightarrow$ $\sqrt{{(a-1)}^2+{(2a-14+2)}^2}$ = $\frac{|4a+3(2a-14)-23|}{5}$ [From (1)]
$\Rightarrow$ 5$\sqrt{a^2-2a+1+{4a}^2-48a+144}$ = $|10a-65|$ $\Rightarrow$ 5$\sqrt{{5a}^2-50a+145}$ = 5$|2a-13|$
$\Rightarrow$ 5a${}^2$ - 50a + 145 = (2a - 13)${}^2$ = 4a${}^2$ - 52a + 169
$\Rightarrow$ a${}^2$ + 2a - 24 = 0 $\Rightarrow$ (a + 6)(a - 4) $\Rightarrow$ a = 4
Case 1 : a = 4 $\Rightarrow$ b = 2(4) - 14 = - 6.$\therefore$ Centre C = (4, - 6)
Radius = CP = $\sqrt{{(4-1)}^2+{(-6+2)}^2}$ = $\sqrt{9+16}$ = 5
The circle equation is $(x-4{)}^2+(y+6{)}^2=5^2$ $\Rightarrow$ $x^2$ + $y^2$ - 8x + 12y + 27 = 0
Case 2 : a = - 6 $\Rightarrow$ b = 2(- 6) - 14 = 26 $\therefore$ Centre C = (-6, - 26)
Radius = CP = $\sqrt{{(-6-1)}^2+{(-26+2)}^2}$ = $\sqrt{49+576}$ = $\sqrt{625}$ = 25
The circle equation is $(x+6{)}^2+(y+26{)}^2={25}^2$ $\Rightarrow$ $x^2$ + $y^2$ + 12x + 52y + 87 = 0
$\therefore$ The circles are $x^2$ + $y^2$ - 8x + 12y + 27 = 0, $x^2$ + $y^2$ + 12x + 52y + 87 = 0