The equation of the circle whose diameter lin | vedclass

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

Question

(A) The center of the circle is the intersection of the diameters $2x + 3y = 3$ and $16x - y = 4$.Multiplying the second equation by $3$,we get $48x -

Vedclass · Accepted Answer

$5(x^2 + y^2) - 3x - 8y = 200$

Vedclass · Answer

(A) The center of the circle is the intersection of the diameters $2x + 3y = 3$ and $16x - y = 4$.Multiplying the second equation by $3$,we get $48x - 3y = 12$.Adding this to the first equation: $(2x + 3y) + (48x - 3y) = 3 + 12$ $\Rightarrow 50x = 15$ $\Rightarrow x = \frac{3}{10}$.Substituting $x = \frac{3}{10}$ into $16x - y = 4$: $16(\frac{3}{10}) - y = 4$ $\Rightarrow \frac{48}{10} - 4 = y$ $\Rightarrow y = \frac{8}{10} = \frac{4}{5}$.So,the center is $(\frac{3}{10}, \frac{4}{5})$.The circle passes through $(4, 6)$,so the radius squared $r^2$ is the distance squared from the center to $(4, 6)$:$r^2 = (4 - \frac{3}{10})^2 + (6 - \frac{4}{5})^2 = (\frac{37}{10})^2 + (\frac{26}{5})^2 = \frac{1369}{100} + \frac{676}{25} = \frac{1369 + 2704}{100} = \frac{4073}{100}$.The equation is $(x - \frac{3}{10})^2 + (y - \frac{4}{5})^2 = \frac{4073}{100}$.Expanding this: $x^2 - \frac{3}{5}x + \frac{9}{100} + y^2 - \frac{8}{5}y + \frac{16}{25} = \frac{4073}{100}$.$x^2 + y^2 - \frac{3}{5}x - \frac{8}{5}y + \frac{9 + 64}{100} = \frac{4073}{100}$ $\Rightarrow x^2 + y^2 - \frac{3}{5}x - \frac{8}{5}y = \frac{4000}{100} = 40$.Multiplying by $5$: $5(x^2 + y^2) - 3x - 8y = 200$.

The equation of the circle whose diameter lines are $2x + 3y = 3$ and $16x - y = 4$ and which passes through the point $(4, 6)$ is

Similar Questions

The equation of the circle whose diameters have the end points $(a, 0)$ and $(0, b)$ is given by:

$A$ circle has its equation in the form $x^2 + y^2 + 2x + 4y + 1 = 0$. Choose the correct coordinates of its centre and the right value of its radius from the following:

The equation of the circle touching the lines $|x-2|+|y-3|=4$ is

The equations of the circle which pass through the origin and make intercepts of lengths $4$ and $8$ on the $x$ and $y$-axes respectively are

Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module