The equation of the circle passing through th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.Since it passes through $(-1, -3)$,we have $1 - 3 + 2g(-1) + 2f(-3) + c =

Vedclass · Accepted Answer

$x^2 + y^2 - 2x + 3y - 3 = 0$

Vedclass · Answer

(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ $(i)$.Since it passes through $(-1, -3)$,we have $1 - 3 + 2g(-1) + 2f(-3) + c = 0 \Rightarrow 10 - 2g - 6f + c = 0$ $(ii)$.Since it passes through $(3, 0)$,we have $9 + 0 + 2g(3) + 0 + c = 0 \Rightarrow 9 + 6g + c = 0$ $(iii)$.The centre of the circle is $C(-g, -f)$. The line $4x + 3y - 12 = 0$ is tangent at $(3, 0)$,so the normal at $(3, 0)$ passes through the centre $(-g, -f)$.The slope of the tangent is $-4/3$,so the slope of the normal is $3/4$.Thus,$\frac{-f - 0}{-g - 3} = \frac{3}{4}$ $\Rightarrow -4f = -3g - 9$ $\Rightarrow 3g - 4f + 9 = 0$ $(iv)$.Solving equations $(ii)$,$(iii)$,and $(iv)$:From $(iii)$,$c = -9 - 6g$. Substituting into $(ii)$: $10 - 2g - 6f - 9 - 6g = 0$ $\Rightarrow 1 - 8g - 6f = 0$ $\Rightarrow 8g + 6f = 1$.We have the system: $3g - 4f = -9$ and $8g + 6f = 1$.Multiplying the first by $3$ and the second by $2$: $9g - 12f = -27$ and $16g + 12f = 2$.Adding them: $25g = -25 \Rightarrow g = -1$.Substituting $g = -1$ into $3g - 4f = -9$: $-3 - 4f = -9$ $\Rightarrow -4f = -6$ $\Rightarrow f = 3/2$.Substituting $g = -1$ into $c = -9 - 6g$: $c = -9 - 6(-1) = -3$.The equation is $x^2 + y^2 - 2x + 3y - 3 = 0$.

The equation of the circle passing through the point $(-1, -3)$ and touching the line $4x + 3y - 12 = 0$ at the point $(3, 0)$ is

Similar Questions

For what value of $\lambda$ does the line $3x - 4y = \lambda$ touch the circle $x^2 + y^2 - 4x - 8y - 5 = 0$?

If a line passing through the origin touches the circle $(x - 4)^2 + (y + 5)^2 = 25$,then its slope is:

If $a > 2b > 0$,then the positive value of $m$ for which $y = mx - b\sqrt{1 + m^2}$ is a common tangent to $x^2 + y^2 = b^2$ and $(x - a)^2 + y^2 = b^2$ is:

At which point on the $y$-axis is the line $x = 0$ a tangent to the circle $x^2 + y^2 - 2x - 6y + 9 = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module