The equation of a circle touching the parabol | vedclass

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(A) The equation of the tangent to the parabola $y = x^2$ at point $P(1, 1)$ is found by differentiating $y = x^2$,giving $\frac{dy}{dx} = 2x$. At $x

Vedclass · Accepted Answer

$x^2 + y^2 - 6x + 4 = 0$

Vedclass · Answer

(A) The equation of the tangent to the parabola $y = x^2$ at point $P(1, 1)$ is found by differentiating $y = x^2$,giving $\frac{dy}{dx} = 2x$. At $x = 1$,the slope $m = 2$. The equation of the tangent is $y - 1 = 2(x - 1)$,which simplifies to $2x - y - 1 = 0$.The equation of a circle touching the line $2x - y - 1 = 0$ at $(1, 1)$ is given by the family of circles equation:$(x - 1)^2 + (y - 1)^2 + \lambda(2x - y - 1) = 0 \quad (i)$Since the circle passes through the point $Q(2, 2)$,we substitute these coordinates into equation $(i)$:$(2 - 1)^2 + (2 - 1)^2 + \lambda(2(2) - 2 - 1) = 0$$1 + 1 + \lambda(4 - 2 - 1) = 0$$2 + \lambda(1) = 0$$\lambda = -2$Substituting $\lambda = -2$ back into equation $(i)$:$(x - 1)^2 + (y - 1)^2 - 2(2x - y - 1) = 0$$x^2 - 2x + 1 + y^2 - 2y + 1 - 4x + 2y + 2 = 0$$x^2 + y^2 - 6x + 4 = 0$

The equation of a circle touching the parabola $y = x^2$ at the point $(1, 1)$ and passing through the point $(2, 2)$ is:

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